Q18. - Algebra 2

Question

Complete parts a-c for each quadratic function.

1. Find y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.
2. Make a table of values that includes the vertex.
3. Use this information to graph the function.

$f\left(x\right)=2x^2-4$

Step-by-Step Solution

Answer

1. The y-intercept is -4, equation of axis of symmetry is $x=0$and x-coordinate of vertex is 0.
2. The table is:

Here, the vertex is $\left(0,-4\right)$

      c. The graph is:

a Step 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 $f\left(x\right)=2x^2-4$

Step 3. Solution.

In the function $f\left(x\right)=2x^2-4$, we have $a=2,b=0,c=-4$

So, the y-intercept is -4.

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

Substitute the values $a=2,b=0$ to get:

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

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

b Step 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 $f\left(x\right)=2x^2-4$ .

From part (a), we have y-intercept is -4, 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 $\left(0,-4\right)$

c Step 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 $f\left(x\right)=2x^2-4$

From part (a) and (b), we have 

Here, the vertex is $\left(0,-4\right)$, the y-intercept is -4, 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.