The quadratic function has vertex and passes through the point . Which equation defines ?

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SAT Nonlinear Functions Question 32 | Free SAT Prep | Aniko

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Question 32·Medium·Nonlinear Functions

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The quadratic function $h$ has vertex $(-2,3)$ and passes through the point $(0,-13)$ . Which equation defines $h$ ?

For quadratic questions that give you a vertex and another point, immediately write the function in vertex form $y=a(x-h)^2+k$ using the vertex $(h,k)$ , then plug the other point in to solve quickly for $a$ . On multiple-choice questions, you can often first eliminate any options with the wrong vertex (wrong sign inside the parentheses or wrong constant term), then use substitution of the given point to choose between the remaining options, which saves time and reduces mistakes.

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Recall vertex form

Think about the vertex form of a quadratic: $y=a(x-h)^2+k$ . How do $h$ and $k$ relate to the vertex $(h,k)$ ?

Use the vertex to narrow down choices

The vertex is $(-2,3)$ . In vertex form $a(x-h)^2+k$ , what should $h$ and $k$ be? Which answer choices show $(x+2)^2$ and end with $+3$ ?

Use the point to determine a

Once you know the form $h(x)=a(x+2)^2+3$ , plug in the point $(0,-13)$ for $x$ and $h(x)$ to solve for $a$ . Then check which option has that value of $a$ .

Desmos Guide

Enter the four candidate functions

Type each option into Desmos as a separate function: $y=-4(x+2)^2+7$ , $y=-2(x+2)^2+3$ , $y=-2(x+2)^2+7$ , and $y=-4(x+2)^2+3$ .

Plot the given points

Add the points $(-2,3)$ and $(0,-13)$ in Desmos by typing them exactly as $( -2, 3 )$ and $( 0, -13 )$ so you can see them on the graph.

Identify the matching graph

Look at which parabola has its vertex exactly at $(-2,3)$ and also passes through the point $(0,-13)$ . The equation of that matching parabola is the one you should choose as your answer.

Step-by-step Explanation

Write the quadratic in vertex form using the vertex

The vertex form of a quadratic is $y=a(x-h)^2+k$ , where $(h,k)$ is the vertex.

Here, the vertex is $(-2,3)$ , so $h=-2$ and $k=3$ . That means the function must look like

h(x)=a(x-(-2))^2+3 = a(x+2)^2+3

So the coefficient $a$ is the only unknown left.

Use the point the graph passes through to find a

We are told the graph passes through $(0,-13)$ , so when $x=0$ , $h(x)=-13$ .

Substitute $x=0$ and $h(x)=-13$ into $h(x)=a(x+2)^2+3$ :

-13 = a(0+2)^2 + 3

This simplifies to $-13 = 4a + 3$ . Now solve for $a$ :

-13 - 3 = 4a \Rightarrow -16 = 4a \Rightarrow a = -4

Write the final equation and match it to a choice

We found $a=-4$ , and we already know the structure is $h(x)=a(x+2)^2+3$ .

Substitute $a=-4$ to get the full equation:

h(x)=-4(x+2)^2+3

This matches choice D.

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