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Stels [109]
3 years ago
10

H(r)=−(r+9) 2 +36h, left parenthesis, r, right parenthesis, equals, minus, left parenthesis, r, plus, 9, right parenthesis, squa

red, plus, 36 1) What are the zeros of the function? Write the smaller rrr first, and the larger rrr second. \text{smaller }r=smaller r=start text, s, m, a, l, l, e, r, space, end text, r, equals \text{larger }r=larger r=start text, l, a, r, g, e, r, space, end text, r, equals 2) What is the vertex of the parabola? ((left parenthesis ,,comma ))right parenthesis
Mathematics
1 answer:
Bond [772]3 years ago
4 0

Answer:

Roots: -15, -9

Vertex: (-9, 36)

Step-by-step explanation:

H(r)=-(r+9)^2+36 is the equation of a parabola in "vertex form."  This form has the answer to the vertex question almost right in front of us.  You may have seen something like this pattern

y=a(x-h)^2+k  as the general vertex form of a parabola's equation. So in your problem, there's some role playing going on.  The role of <em>h</em> is being played by -9, the role of <em>k</em> is being played by 36.  (And <em>x</em> is being played by <em>r</em>.)

The vertex is the point (<em>h</em>, <em>k</em>) = (-9, 36) in your problem.  By the way, the vertex occurs where the quantity (<em>r</em> + 9) is equal to 0, so that's pretty simple to find; it's -9.  If <em>r</em> is replaced with -9, the value of the function is 0 + 36 = 36, so 36 is the second coordinate of the vertex.

Now for the zeros.  "Zeros" means to find the values of <em>r</em> that make the function's value equal to 0.

-(r+9)^2+36=0 \\ -(r+9)^2=-36 \\ (r+9)^2=36 \\ r+9=\pm 6

There are two possibilities:

r+9=6 \\ r=-3

or

r+9=-6 \\ r=-15

The zeros are -15 and -3 (the smaller one is -15).

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see the attached figure to better understand the problem

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