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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?​

User Secmask
by
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1 Answer

7 votes

Answer:

1) The solutions are:

smaller r = -8

larger r = -1

2) The vertex is
((-9)/(2),(-49)/(4)).

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
-(b)/(2a).

We know from the standard equation of a parabola that


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

So,


r=-(9)/(2\cdot 1) = -(9)/(2)

And the h-coordinate is


h(-(9)/(2))=(-(9)/(2))^2+9(-(9)/(2))+8\\\\=(9^2)/(2^2)-(81)/(2)+8\\\\=(9^2)/(4)-(162)/(4)+(32)/(4)\\\\=(-49)/(4)

The vertex is
((-9)/(2),(-49)/(4)).

User Kary
by
4.9k points