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Q17.

Expert-verifiedFound in: Page 291

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.**

$f\left(x\right)={x}^{2}-9$

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

Here, the vertex is $\left(0,-9\right)$

c. The graph is:

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
is $\frac{-b}{2a}$*x*-coordinate of vertex

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

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:

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

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

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
is $\frac{-b}{2a}$*x*-coordinate of vertex

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.

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.

Here, the vertex is $(0,-9)$

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
is $\frac{-b}{2a}$*x*-coordinate of vertex

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 $x=0$.

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.

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