Question

h(r) = (r + 1)(r +8)
1) What are the zeros of the function?
Write the smaller r first, and the larger r second.
smaller r =
larger r =
2) What is the vertex of the parabola?​

Answer 1

Answer:

1) The solutions are:

smaller r =  -8

larger r = -1

2) The vertex is $(\frac{-9}{2},\frac{-49}{4})$.

Step-by-step explanation:

1) A zero of a function is an x-value that makes the function value 0.

To find the zeros of the function, first, we see that the linear factors of $h(r)$ are

$(r + 1)$ and $(r +8)$.

If we set $h(r)=0$ and solve for x, we get

$\left(r\:+\:1\right)\left(r\:+8\right)=\:0$

Using the Zero factor principle, If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

$r+1=0:\quad r=-1\\r+8=0:\quad r=-8$

The solutions are:

smaller r =  -8

larger r = -1

2) The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a parabola.

To find the vertex of the function $h(r) = (r + 1)(r +8)$, first we need to find the standard equation of a parabola, which is,

$y=ax^2+bx+c$

$\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd\\\\a=r,\:b=1,\:c=r,\:d=8\\\\rr+8r+1\cdot \:r+1\cdot \:8\\\\r^2+9r+8$

Next, we find the r-coordinate of the vertex with the formula $-\frac{b}{2a}$.

We know from the standard equation of a parabola that

$a=1\\b=9\\c=8$

So,

$r=-\frac{9}{2\cdot 1} = -\frac{9}{2}$

And the h-coordinate is

$h(-\frac{9}{2})=(-\frac{9}{2})^2+9(-\frac{9}{2})+8\\\\=\frac{9^2}{2^2}-\frac{81}{2}+8\\\\=\frac{9^2}{4}-\frac{162}{4}+\frac{32}{4}\\\\=\frac{-49}{4}$

The vertex is $(\frac{-9}{2},\frac{-49}{4})$.