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Expert-verified Found in: Page 291 ### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903 # Complete parts a-c for each quadratic function.Find y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.Make a table of values that includes the vertex.Use this information to graph the function.$f\left(x\right)={x}^{2}-9$

1. The y-intercept is -9, equation of axis of symmetry is $x=0$and x-coordinate of vertex is 0.
2. The table is:

Here, the vertex is $\left(0,-9\right)$ c. The graph is: See the step by step solution

## aStep 1. Use the concept.

Consider the function $f\left(x\right)=a{x}^{2}+bx+c,a\ne 0$

• The y-intercept is $f\left(x\right)=a{\left(0\right)}^{2}+b\left(0\right)+c\text{or}c$
• The equation of axis of symmetry is $x=\frac{-b}{2a}$
• The x-coordinate of vertex is $\frac{-b}{2a}$

## Step 2. Given Information.

The given function is $f\left(x\right)={x}^{2}-9$

## Step 3. Solution.

In the function $f\left(x\right)={x}^{2}-9$, we have $a=1,b=0,c=-9$

So, the y-intercept is -9.

The equation of axis of symmetry is given by $x=\frac{-b}{2a}$

Substitute the values $a=1,b=0$ to get:

$\begin{array}{l}x=\frac{-0}{2\left(1\right)}...............\left[a=1,b=0\right]\\ ⇒x=0\end{array}$

So, the equation of axis of symmetry is $x=0$. Therefore, the x-coordinate of vertex is 0.

## bStep 1. Use the concept.

Consider the function $f\left(x\right)=a{x}^{2}+bx+c,a\ne 0$

• The y-intercept is $f\left(x\right)=a{\left(0\right)}^{2}+b\left(0\right)+c\text{or}c$
• The equation of axis of symmetry is $x=\frac{-b}{2a}$
• The x-coordinate of vertex is $\frac{-b}{2a}$

## Step 2. Given Information.

The given function is $f\left(x\right)={x}^{2}-9$.

From part (a), we have y-intercept is -9, equation of axis of symmetry is $x=0$ and x-coordinate of vertex is 0.

## Step 3. Discussion.

Choose some values for x that are less than 0 and some that are greater than 0. This ensures that points on each side of the axis of symmetry are graphed.

## Step 4. Table. Here, the vertex is $\left(0,-9\right)$

## cStep 1. Use the concept.

The graph of $f\left(x\right)=a{x}^{2}+bx+c,a\ne 0$

• The y-intercept is $f\left(x\right)=a{\left(0\right)}^{2}+b\left(0\right)+c\text{or}c$
• The equation of axis of symmetry is $x=\frac{-b}{2a}$
• The x-coordinate of vertex is $\frac{-b}{2a}$

## Step 2. Given Information.

The given function is $f\left(x\right)={x}^{2}-9$

From part (a) and (b), we have Here, the vertex is $\left(0,-9\right)$, the y-intercept is -9, equation of axis of symmetry is $x=0$.

## Step 3. Solution.

Graph the vertex and y-intercept. Then graph the points from your table connecting them and the y-intercept with a smooth curve. As a check, draw the axis of symmetry, $x=0$, as a green line. The graph of the function should be symmetrical about this line.  ### Want to see more solutions like these? 