Question

Graph the function f(x) = – 4(x + 4)2 square + 7.
Plot the vertex. Then plot another point on the parabola. If you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Answer 1

• Vertex at (-4,7)

Graph attached

Answer 2

Answer:

vertex = (-4, 7)

(-3, 3) and (-2, -9)

(-6, -9) and (-5, 3)

Step-by-step explanation:

Given function:  $f(x) = -4(x + 4)^2 + 7$

Vertex form of a quadratic function:  $f(x)=a(x-h)^2+k$
(where (h, k) is the vertex)

From inspection, we can see that the given function is in vertex form.

h = -4 and k = 7

Therefore, the vertex is at (-4, 7)

As $a$ is negative, the parabola will open downwards.

When graphing functions, it is useful to determine the axis intercepts.

The curve will intercept the y-axis when x = 0.

Substituting x = 0 into the function:

$f(0) = -4(0 + 4)^2 + 7=-57$

Therefore, the y-intercept is at (0, -57)

**cannot plot this on the given graph area as it is out of range**

The curve will intercept the x-axis when y = 0.

Setting the function to 0 and solving for x:

$-4(x + 4)^2 + 7=0$

$\implies -4(x + 4)^2=-7$

$\implies (x + 4)^2=\dfrac74$

$\implies x+4=\pm\sqrt{\dfrac74}$

$\implies x=-4\pm\dfrac{\sqrt{7}}{2}$

Therefore, the x-intercepts are at (-2.7, 0) and (-5.3, 0) to 1 dp.

Finally, input values of x either sides of the x-intercepts for further plot points:

(-3, 3) and (-2, -9)

(-6, -9) and (-5, 3)