Topics
7. Quadratic Functions
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Algebra 1

7.04 Quadratic functions in standard form

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Practice

Quadratic functions in standard form

Exploration

Move the a, b, and c sliders to transform the graph.

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  1. What happens to the graph as the value of a changes?
  2. What happens to the graph as the value of b changes?
  3. What happens to the graph as the value of c changes?

The standard form of a quadratic equation, where a, b, and c are real numbers is:

\displaystyle y=ax^2+bx+c
\bm{a}
scale factor
\bm{b}
linear coefficient
\bm{c}
y-value of the y-intercept
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If a>0, the parabola opens upwards. The vertex will be a minimum point.
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If a<0, the parabola opens downwards. The vertex will be a maximum point.

The x-intercepts of the function, also known as roots or zeros, are the points where the graph crosses the x-axis. This is where y=0.

Other key features of a quadratic function help describe its graph and behavior.

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\, \\The domain is all real numbers, -\infty \leq x \leq \infty. The range depends on the vertex's y-coordinate (k) and opening direction: y \geq k for upwards, y \leq k for downwards.

The function increases on one side of the vertex and decreases on the other. The end behavior sees both ends of the graph pointing up or down.

The standard form of a quadratic equation allows us to quickly identify the y-intercept and whether the parabola opens up or down.

The coordinates of the vertex are: \left(-\dfrac{b}{2a},\,f\left(-\dfrac{b}{2a}\right)\right)

We can substitute the x-coordinate of the vertex into the original equation to find the y-coordinate of the vertex.

We can also see that the axis of symmetry is the line:

x=-\dfrac{b}{2a}

The axis of symmetry always passes through the vertex.

To graph a quadratic function from standard form, we can follow these steps:

  1. Determine if the parabola opens upwards (a > 0) or downwards (a < 0).

  2. Find the y-intercept by identifying the value of c. Plot the point (0, c).

  3. Find the axis of symmetry using x = -\dfrac{b}{2a}. This is the x-coordinate of the vertex.

  4. Find the y-coordinate of the vertex by substituting the x-value from the axis of symmetry into the function. Plot the vertex.

  5. Find the x-intercepts by setting y=0 and solving for x using the quadratic formula or factoring. Plot these points if they are real numbers.

  6. Use the axis of symmetry to find additional points. For example, reflect the y-intercept across the axis of symmetry.

  7. Draw a smooth curve through the plotted points to form the parabola.

Examples

Example 1

For the quadratic function y=3x^2-6x+8, find the key features and graph the function.

a

Determine the direction of opening and the y-intercept.

Worked Solution
Create a strategy

The standard form y=ax^2+bx+c gives us two key features directly. The sign of the coefficient a tells us the direction of opening, and the constant c is the y-value of the y-intercept.

Apply the idea

In the function y=3x^2-6x+8, we have a=3, b=-6, and c=8.

Since a=3 is positive, the parabola opens upwards.

The value of c is 8, so the y-intercept is at (0, 8).

b

Find the axis of symmetry and the vertex.

Worked Solution
Create a strategy

We can find the axis of symmetry using the formula x = -\dfrac{b}{2a}. The x-value of the vertex is on the axis of symmetry. We can substitute this value back into the function to find the y-value of the vertex.

Apply the idea

Find the axis of symmetry:

\displaystyle x\displaystyle =\displaystyle -\dfrac{-6}{2(3)}Substitute a=3 and b=-6
\displaystyle x\displaystyle =\displaystyle \dfrac{6}{6}Simplify
\displaystyle x\displaystyle =\displaystyle 1The axis of symmetry is x=1.

Now, find the vertex by substituting x=1 into the function:

\displaystyle y\displaystyle =\displaystyle 3(1)^2 - 6(1) + 8Substitute x=1
\displaystyle y\displaystyle =\displaystyle 3 - 6 + 8Simplify
\displaystyle y\displaystyle =\displaystyle 5The vertex is at (1, 5).
c

Graph the function.

Worked Solution
Create a strategy

We will plot the key points we have found: the vertex (1, 5) and the y-intercept (0, 8).

We can find another point by reflecting the y-intercept across the axis of symmetry, x=1, to get the point (2, 8).

Apply the idea
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Example 2

Find the x-intercepts of the function y = -x^2 + 4x + 5 using the quadratic formula.

Worked Solution
Create a strategy

To find the x-intercepts, we set y=0. The function is in standard form, so we can identify a, b, and c and substitute them into the quadratic formula: x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Apply the idea

For the equation 0 = -x^2 + 4x + 5, we have a = -1, b = 4, and c = 5.

Substitute these values into the quadratic formula:

\displaystyle x\displaystyle =\displaystyle \dfrac{-(4) \pm \sqrt{(4)^2 - 4(-1)(5)}}{2(-1)}Substitute a, b, and c
\displaystyle x\displaystyle =\displaystyle \dfrac{-4 \pm \sqrt{16 + 20}}{-2}Simplify inside the radical
\displaystyle x\displaystyle =\displaystyle \dfrac{-4 \pm \sqrt{36}}{-2}
\displaystyle x\displaystyle =\displaystyle \dfrac{-4 \pm 6}{-2}Evaluate the square root

Now, we find the two separate solutions:

x = \dfrac{-4 + 6}{-2} = \dfrac{2}{-2} = -1

x = \dfrac{-4 - 6}{-2} = \dfrac{-10}{-2} = 5

The x-intercepts are at (-1, 0) and (5, 0).

Reflect and check

We can check our answer by factoring the quadratic. -x^2 + 4x + 5 can be factored as -(x-5)(x+1). Setting this to zero gives roots of x=5 and x=-1, which confirms our answer.

Example 3

Naomi is playing a game of Kapucha Toli, where to start a play, a ball is thrown into the air.

Naomi throws a ball into the air from a height of 6 feet, and the maximum height the ball reaches is 12.25 feet after 1.25 seconds.

a

Sketch a graph to model the height of the ball over time.

Worked Solution
Create a strategy

To sketch a graph, we'll use key points found by using the given information. We'll also use the units which are given, being feet and seconds.

We will let x represent the time since the ball was tossed in seconds.

We will let y represent the height of the ball in feet.

Apply the idea

It is given that the ball is thrown from a height of 6 feet. This means that at 0 seconds, the height of the ball is 6 feet. So our y-intercept is \left(0,\,6 \right).

We are told the maximum height of the ball is at 12.25 feet after 1.25 seconds. The maximum height will occur at the vertex of the graph, so the vertex is \left(1.25,\,12.25\right). This also means that our axis of symmetry is x=1.25.

We can use the axis of symmetry to determine a second point on the graph, the point across the axis of symmetry from the y-intercept. The point is \left(2.5,\,6 \right).

We can sketch our graph by plotting the y-intercept, the vertex, and the point found with our axis of symmetry.

Now we need to identify an appropriate scale.

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\text{Time in seconds }(x)
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\text{Height in feet }(y)

We know that our graph will not go above y=12.25\text{ feet} and that any part of the graph that goes below the x-axis will not be viable.

Graphing -1 \leq y \leq 13 with an interval of 1 will show the full picture.

We know that time starts at x=0 and the ball is on the way back down at x=2.5. So graphing 0 \leq x \leq 4 with an interval of 1 or 0.5 should be sufficient.

This is an appropriate way to label the axes.

Now, we can graph the height of the ball over time.

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\text{Time in seconds }(x)
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\text{Height in feet }(y)
b

Predict when the ball will be 3 feet above the ground.

Worked Solution
Create a strategy

We can use the sketch of our graph to predict when the ball will be at 3 feet.

Apply the idea

We can draw a horizontal line from y=3 across until we reach the graph.

After that, we can draw vertical line until we reach the x-axis to determine after how many seconds the ball is at 3 feet.

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\text{Time in seconds }(x)
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\text{Height in feet }(y)

We hit the x-axis around x=2.7. Therefore, the ball is 3\text{ ft} above the ground after about 2.7 seconds.

Reflect and check

When reading from a graph, we often have to estimate. Any prediction between 2.6 and 2.9 would be reasonable in this case.

c

Write a quadratic equation in standard form to model the situation.

Worked Solution
Create a strategy

To write the equation, we can use the key points and the graph we've sketched in previous parts.

Apply the idea

Since we know 3 points, we can use the standard form and substitution to solve for a, b, and c for our standard form quadratic equation, which is of the form y=ax^2+bx+c.

We know that c represents the y-value of the y-intercept, which is \left(0,\,6 \right):y=ax^2+bx+6

Now, we can substitute our other two points to solve for a and b.

Next, we can substitute in \left(1.25,\,12.25\right).

\displaystyle y\displaystyle =\displaystyle ax^2+bx+6Standard form of a quadratic with c=6
\displaystyle 12.25\displaystyle =\displaystyle a(1.25)^2+b(1.25)+6Substitute \left(1.25,\,12.25\right)
\displaystyle 12.25\displaystyle =\displaystyle 1.5625a + 1.25b + 6Evaluate the exponent
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25bSubtract 6 from both sides

Since we have two unknowns, we'll have to use our final point to create a second equation.

We'll now substitute in \left(2.5,\,6 \right).

\displaystyle y\displaystyle =\displaystyle ax^2+bx+6Standard form of a quadratic with c=6
\displaystyle 6\displaystyle =\displaystyle a(2.5)^2+b(2.5)+6Substitute \left(2.5,\,6\right)
\displaystyle 6\displaystyle =\displaystyle 6.25a + 2.5b + 6Evaluate the exponent
\displaystyle 0\displaystyle =\displaystyle 6.25a + 2.5bSubtract 6 from both sides

Now, we have a system of two equations with two unknowns. We can solve this using substitution. Let's first isolate b in our second equation.

\displaystyle 0\displaystyle =\displaystyle 6.25a + 2.5bSecond equation
\displaystyle -6.25a\displaystyle =\displaystyle 2.5bSubtract 6.25a from both sides
\displaystyle \dfrac{-6.25a}{2.5}\displaystyle =\displaystyle bDivide by 2.5 on both sides
\displaystyle -2.5a\displaystyle =\displaystyle bEvaluate the division

Now, we can use this in our first equation, letting b = -2.5a.

\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25bFirst equation
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a + 1.25(-2.5a)Substitute b = -2.5a
\displaystyle 6.25\displaystyle =\displaystyle 1.5625a - 3.125aEvaluate the multiplication
\displaystyle 6.25\displaystyle =\displaystyle -1.5625aCombine like terms
\displaystyle -4\displaystyle =\displaystyle aDivide both sides by -1.5625

Therefore a=-4. Finally, we can use b = -2.5a to solve for b.

\displaystyle b\displaystyle =\displaystyle -2.5a
\displaystyle b\displaystyle =\displaystyle -2.5(-4)Substitute a = -4
\displaystyle b\displaystyle =\displaystyle 10Evaluate the multiplication

Therefore b=10. Now it's time to piece it all together. Since a=-4, b=10, and c=6, we know that our equation in standard form is: y = -4x^2 + 10x + 6

Reflect and check

An alternative and simpler solution is to use the vertex to write it in vertex form and then use the intercept to solve for a.

Vertex form is y=a(x-h)^2+k. The vertex is \left(1.25,\,12.25\right), this gives us: y=a(x-1.25)^2+12.25

We can then substitute in the point \left(0,\,6\right) and solve for a.

\displaystyle y\displaystyle =\displaystyle a(x-1.25)^2+12.25Vertex form of a quadratic with vertex \left(1.25,\,12.25\right)
\displaystyle 6\displaystyle =\displaystyle a(0-1.25)^2+12.25Substitute in \left(0,\,6\right)
\displaystyle 6\displaystyle =\displaystyle 1.5625a+12.25Evaluate the parentheses
\displaystyle -6.25\displaystyle =\displaystyle 1.5625aSubtract 12.25 from both sides
\displaystyle \dfrac{-6.25}{1.5625}\displaystyle =\displaystyle aDivide by 1.5625 on both sides
\displaystyle -4\displaystyle =\displaystyle aEvaluate the division

So, now we have the equation: y=-4(x-1.25)^2+12.25

Now, we need to convert to standard form:

\displaystyle y\displaystyle =\displaystyle -4(x-1.25)^2+12.25Equation in vertex form
\displaystyle y\displaystyle =\displaystyle -4(x^2-2.5x+1.5625)+12.25Expand the binomial
\displaystyle y\displaystyle =\displaystyle -4x^2+10x-6.25+12.25Distributive property
\displaystyle y\displaystyle =\displaystyle -4x^2+10x+6Combine like terms

We get the same answer of: y=-4x^2+10x+6.

Example 4

Write the standard form equation for the function shown in the graph.

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Worked Solution
Create a strategy

Use the vertex to first write the equation in vertex form. Then use another point on the parabola to solve for a, and finally convert to standard form.

Apply the idea

Vertex form is y=a(x-h)^2+k. The vertex is \left(8, -2\right), so this gives us: y=a(x-8)^2-2

We can then substitute in the point \left(4,\,6\right) and solve for a.

\displaystyle y\displaystyle =\displaystyle a(x-8)^2-2Vertex form equation
\displaystyle 6\displaystyle =\displaystyle a(4-8)^2-2Substitute x=4 and y=6
\displaystyle 6\displaystyle =\displaystyle a(-4)^2-2Evaluate the subtraction
\displaystyle 6\displaystyle =\displaystyle 16a - 2Evaluate the exponent
\displaystyle 8\displaystyle =\displaystyle 16aAdd 2 to both sides
\displaystyle \dfrac{8}{16}\displaystyle =\displaystyle aDivide by 16 on both sides
\displaystyle \dfrac{1}{2}\displaystyle =\displaystyle aSimplify the fraction

Substituting a=\dfrac{1}{2} we get the equation: y=\dfrac{1}{2}(x-8)^2 - 2

Now, we need to convert to standard form:

\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}(x-8)^2 - 2Equation in vertex form
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}(x^2 - 16x + 64) - 2Expand the binomial
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}x^2 - 8x + 32 - 2Distributive property
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}x^2 - 8x + 30Combine like terms

The standard form equation for the function is y=\dfrac{1}{2}x^2 - 8x + 30.

Reflect and check

Alternatively, we could have used the zeros and factored form.

Using the zeros of 6 and 10, we can write the equation in factored form, y=a(x - x_1)(x - x_2). This gives us:y=a(x - 6)(x - 10)

We can then substitute in the point \left(4,\,6\right) and solve for a.

\displaystyle y\displaystyle =\displaystyle a(x - 6)(x - 10)Factored form equation
\displaystyle 6\displaystyle =\displaystyle a(4 - 6)(4 - 10)Substitute x=4 and y=6
\displaystyle 6\displaystyle =\displaystyle a(-2)(-6)Evaluate the subtraction
\displaystyle 6\displaystyle =\displaystyle 12aEvaluate the multiplication
\displaystyle \dfrac{6}{12}\displaystyle =\displaystyle aDivide by 12 on both sides
\displaystyle \dfrac{1}{2}\displaystyle =\displaystyle aSimplify the fraction

So, now we have the equation: y=\dfrac{1}{2}(x-6)(x-10)

Now, we need to convert to standard form:

\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}(x-6)(x-10)Equation in factored form
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}(x^2 - 10x - 6x + 60)Multiply the binomials
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}(x^2 - 16x + 60)Combine like terms
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}x^2 - 8x + 30Distributive property

Example 5

A whale's jump is modeled by a quadratic function. The whale leaves the water at 3 seconds and reenters at 6.5 seconds. It reaches a maximum height of 49 feet at 4.75 seconds.

A humpback whale breaching.
a

Determine the equation in standard form that models the whale’s jump.

Worked Solution
Create a strategy

If we think of the water level as the x-axis, then the moments where the whale exits and reenters the water would represent the x-intercepts.

The maximum height is the vertex of the parabola formed by the whale's jump path.

Apply the idea

Using the x-intercepts of 3 and 6.5, we can create the following equation, where x represents the time in seconds and y represents the height of the jump in feet:

y=a(x-3)(x-6.5)

Next, we can substitute the values of the maximum point, which occurs at (4.75,\, 49), to find the value of a.

\displaystyle y\displaystyle =\displaystyle a(x-3)(x-6.5)Equation for whale's path
\displaystyle 49\displaystyle =\displaystyle a(4.75-3)(4.75-6.5)Substitute y=49 and x=4.75
\displaystyle 49\displaystyle =\displaystyle a(1.75)(-1.75)Evaluate the subtraction
\displaystyle 49\displaystyle =\displaystyle -3.0625aEvaluate the multiplication
\displaystyle -16\displaystyle =\displaystyle aDivision property of equality

Now, we know the factored form of the equation that models the whale's jump:

y=-16(x-3)(x-6.5)

The last step is to get it into standard form. We can do this by multiplying all the factors together.

\displaystyle y\displaystyle =\displaystyle -16(x-3)(x-6.5)Equation for whale's path
\displaystyle y\displaystyle =\displaystyle -16(x^2 - 9.5x + 19.5)Combine like terms after multiplying the binomials
\displaystyle =\displaystyle -16x^2 + 152x - 312Distributive property

The equation in standard form that models the whale's jump is y=-16x^2+152x-312.

Reflect and check

Notice that the parabola formed by the whale opens downward. If we did not have another point to help us find the value of a, our parabola would have been facing upward. Using the vertex helped us find a negative value for a, which is what made the parabola face downward.

b

Interpret the domain, range, and intervals of increase and decrease in the context of the whale's jump.

Worked Solution
Create a strategy

We need to consider what the x and y values represent in this real-world scenario. The domain and range must be limited to values that make sense for a whale jumping out of the water.

Apply the idea

  • Domain: The domain represents the time the whale is in the air. The jump starts at x=3 and ends at x=6.5. Therefore, the viable domain for the jump is 3 \leq x \leq 6.5 seconds.

  • Range: The range represents the whale's height above the water. The whale starts at a height of 0 feet, reaches a maximum of 49 feet, and returns to 0. The range is 0 \leq y \leq 49 feet.

  • Interval of Increase: The interval of increase is when the whale's height is increasing from the moment it leaves the water until it reaches its peak. This occurs on the time interval 3 < x < 4.75 seconds.

  • Interval of Decrease: The interval of decrease is when the whale's height is decreasing from its peak until it reenters the water. This occurs on the time interval 4.75 < x < 6.5 seconds.

Idea summary

The standard form of a quadratic equation is y=ax^2+bx+c.

  • The coefficient a determines the direction of opening. If a>0, the parabola opens up. If a<0, it opens down.

  • The constant c gives the y-intercept at (0, c).

  • The axis of symmetry is a vertical line x = -\dfrac{b}{2a}.

  • The vertex has an x-coordinate of -\dfrac{b}{2a}. The y-coordinate is found by substituting this x-value back into the function.

  • The x-intercepts (or roots) are the point or points where the parabola intersects the x-axis.

  • The domain of a quadratic function is all real numbers. The range depends on the vertex and the direction of opening.

  • The function increases on one side of the vertex and decreases on the other.

Outcomes

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2b

Given an equation or graph, determine key characteristics of a quadratic function including x-intercepts (zeros), y-intercept, vertex (maximum or minimum), and domain and range (including when restricted by context); interpret key characteristics as related to contextual situations, where applicable.

A.F.2c

Graph a quadratic function, f(x), in two variables using a variety of strategies, including transformations f(x) + k and kf(x), where k is limited to rational values.

A.F.2d

Make connections between the algebraic (standard and factored forms) and graphical representation of a quadratic function.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

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