If the quadratic function is graphed in the -plane, what is the -coordinate of the vertex of the graph?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 197 | Free SAT Prep | Aniko

00:00

Question 197·Medium·Nonlinear Functions

0 / 234

If the quadratic function

h(x)=x^2+4x-21

is graphed in the $xy$ -plane, what is the $y$ -coordinate of the vertex of the graph?

For vertex questions where the quadratic is in standard form $ax^2+bx+c$ , quickly use the vertex formula $x=-\dfrac{b}{2a}$ to find the $x$ -coordinate, then substitute that value back into the function to get the $y$ -coordinate. Double-check signs carefully when plugging in (especially with negative $x$ -values) to avoid simple arithmetic errors, and remember that an upward-opening parabola ( $a>0$ ) has a minimum $y$ -value while a downward-opening one ( $a<0$ ) has a maximum.

Hints

Click me!

Recall how to find a vertex

For a quadratic in the form $ax^2+bx+c$ , think about the formula that gives the $x$ -coordinate of the vertex in terms of $a$ and $b$ .

Find the x-value first

Use $x=-\dfrac{b}{2a}$ with $a=1$ and $b=4$ to find the $x$ -coordinate of the vertex for this quadratic.

Evaluate the function carefully

Once you have the $x$ -coordinate, substitute it back into $h(x)=x^2+4x-21$ . Be very careful with squaring the negative number and with the signs in each term.

Desmos Guide

Graph the quadratic

In Desmos, type y = x^2 + 4x - 21 to graph the parabola in the coordinate plane.

Locate the vertex

Tap or click on the lowest point of the parabola (its minimum). Desmos will show the coordinates of this vertex; read off the $y$ -value shown there—that is the $y$ -coordinate of the vertex.

Step-by-step Explanation

Identify the coefficients of the quadratic

The function is

h(x)=x^2+4x-21.

This is in the standard form $ax^2+bx+c$ , so:

$a=1$
$b=4$
$c=-21$ .

Find the x-coordinate of the vertex

For a quadratic $y=ax^2+bx+c$ , the $x$ -coordinate of the vertex is given by

x=-\frac{b}{2a}.

Substitute $a=1$ and $b=4$ :

x=-\frac{4}{2\cdot 1}=-\frac{4}{2}=-2.

So the vertex occurs at $x=-2$ .

Substitute to find the y-coordinate of the vertex

Now plug $x=-2$ into $h(x)$ to find the $y$ -coordinate:

\begin{aligned} h(-2) &= (-2)^2 + 4(-2) - 21 \\ &= 4 - 8 - 21 \\ &= -4 - 21 \\ &= -25. \end{aligned}

So, the $y$ -coordinate of the vertex is $\boxed{-25}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation