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JulijaS [17]
3 years ago
10

NEED HELP ASAP!

Mathematics
1 answer:
Brrunno [24]3 years ago
8 0
complette the square to get vertex form or y=a(x-h)^2+k
(h,k) is vertex
1. group x terms, so for y=ax^2+bx+c, do y=(ax^2+bx)+c <span> <span> </span> 2, factor out the leading coefinet (constant in front of the x^2 term), basicallly factor out a </span><span> <span> </span> 3. take 1/2 of the linear coefient (number in front of the x), and square it ,then add negative and positive of it inside parnthases </span><span> <span> </span> 4. complete the squre and expand </span>


so
y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
vertex is (8,-3)
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Step-by-step explanation:

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1) f(x) = 2x + 4, g(x) = 4x2 + 1; Find (g ∘ f)(0).
Sholpan [36]

Answer:

<h2>(g \: \circ \: f)(0) = 17</h2>

Step-by-step explanation:

f(x) = 2x + 4

g(x) = 4x² + 1

In order to find (g ∘ f)(0) we must first find

(g ° f )(x)

To find (g ° f )(x) substitute f(x) into g(x) that's for every x in g(x) replace it with f(x)

That's

<h3>(g \: \circ \: f)(x) = 4( ({2x + 4})^{2} ) + 1 \\  = 4(4 {x}^{2}  + 16x + 16) + 1 \\  =  {16x}^{2}  + 64x + 16 + 1</h3>

We have

<h3>(g \: \circ \: f)(x) =  {16x}^{2}  + 64x + 17 \\</h3>

Now to find (g ∘ f)(0) substitute the value of x that's 0 into (g ∘ f)(0)

We have

<h3>(g \: \circ \: f)(0) = 16( {0})^{2}  + 64(0) + 17 \\</h3>

We have the final answer as

<h3>(g \: \circ \: f)(0) = 17</h3>

Hope this helps you

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