The quadratic function has axis of symmetry . The graph of passes through and . What is the -coordinate of the vertex of the graph?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 192 | Free SAT Prep | Aniko

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Question 192·Hard·Nonlinear Functions

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The quadratic function $f$ has axis of symmetry $x=1$ . The graph of $y=f(x)$ passes through $(0,0)$ and $(4,27)$ . What is the $y$ -coordinate of the vertex of the graph?

For quadratic function questions involving the vertex and symmetry, first use the given axis of symmetry to fix the vertex’s $x$ -coordinate. If you see one $x$ -intercept, reflect it across the axis of symmetry to find the other intercept, then write the function in factored form $a(x-r_1)(x-r_2)$ . Use any additional point to solve for $a$ , and finally plug the vertex’s $x$ -coordinate into your function to get the vertex’s $y$ -value. This approach avoids heavy algebra and keeps the work organized and quick.

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Start with the axis of symmetry

The axis of symmetry is the vertical line $x=1$ . What does that tell you about the $x$ -coordinate of the vertex?

Use symmetry with the point (0,0)

If $(0,0)$ is on the graph and the parabola is symmetric about $x=1$ , what point must also be on the graph on the other side of $x=1$ with the same $y$ -value?

Write a factored form and use the other point

Once you know the two $x$ -intercepts, you can write $f(x)$ as $a(x-r_1)(x-r_2)$ . Then plug in the point $(4,27)$ to find $a$ , and finally evaluate $f(1)$ to get the vertex’s $y$ -coordinate.

Desmos Guide

Model the quadratic with its roots

In Desmos, type a on a new line to create a slider for $a$ . Then on the next line, enter f(x) = a*x*(x-2) to represent a parabola with $x$ -intercepts at $0$ and $2$ and an adjustable vertical stretch.

Use the point (4,27) to determine a

On another line, type a*4*(4-2) so Desmos shows the value of this expression. Adjust the slider for $a$ until this expression equals $27$ . The value of $a$ at that moment is the stretch factor that makes the graph pass through $(4,27)$ .

Read the vertex’s y-value from f(1)

Finally, on a new line, type f(1) to have Desmos compute the corresponding $y$ -value when $x=1$ . That output is the $y$ -coordinate of the vertex of the parabola.

Step-by-step Explanation

Use the axis of symmetry to locate the vertex horizontally

For any quadratic in the $xy$ -plane, the vertex lies on the axis of symmetry. The axis of symmetry is given as $x=1$ , so the $x$ -coordinate of the vertex must be $1$ . We still need the corresponding $y$ -coordinate $f(1)$ .

Use symmetry to find both x-intercepts and write a factored form

We know the graph passes through $(0,0)$ , so $f(0)=0$ . That means $x=0$ is a root, so $x$ is a factor of $f(x)$ .

Because the axis of symmetry is $x=1$ , any point on the graph must have a mirror image across the line $x=1$ with the same $y$ -value. The point $(0,0)$ is $1$ unit to the left of $x=1$ , so its reflection is $1$ unit to the right, at $(2,0)$ . Thus $x=2$ is also a root.

So $f(x)$ has roots $0$ and $2$ , and we can write it in factored form as

f(x)=a x(x-2)

for some constant $a$ (the vertical stretch factor).

Use the point (4,27) to solve for the stretch factor a

We are told that the graph passes through $(4,27)$ , so $f(4)=27$ .

Substitute $x=4$ into $f(x)=a x(x-2)$ :

f(4)=a\cdot 4\cdot (4-2)=a\cdot 4\cdot 2=8a

Set this equal to $27$ and solve for $a$ :

8a=27 \Rightarrow a=\dfrac{27}{8}

So the quadratic is $f(x)=\dfrac{27}{8}x(x-2)$ .

Find the y-coordinate of the vertex by evaluating f(1)

We already know the vertex has $x$ -coordinate $1$ , so its $y$ -coordinate is $f(1)$ .

Substitute $x=1$ into $f(x)=\dfrac{27}{8}x(x-2)$ :

f(1)=\dfrac{27}{8}\cdot 1\cdot (1-2)=\dfrac{27}{8}\cdot(-1)=-\dfrac{27}{8}

Therefore, the $y$ -coordinate of the vertex is $-\dfrac{27}{8}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation