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gulaghasi [49]
3 years ago
15

Need help with this question. I will give Brainliest!

Mathematics
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

See below

Step-by-step explanation:

a) The vertex is (0,2) so the standard form equation is (x-0)^2=4p(y-2). Since p=1/(4a) and a=1/8, then p=1/(4(1/8))=1/(1/2)=2. This makes the equation now x^2=8(y-2). Given the focus point for a vertical parabola is (h,k+p) then the focus point is (0,2+2) which is (0,4)

b) They are both 4 units away

c) I can't solve

d) I can't solve

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I would believe its 10
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3 years ago
What must be a factor of the polynomial function fx) graphed on the coordinate plane below?
Reptile [31]

See in the explanation

<h2>Explanation:</h2>

Hello! Recall that you need to write complete questions in order to get good and precise answers. However, I'll try to explain this problem in a general way. The definition of polynomial functions states:

A \ \mathbf{polynomial \ function} \ of \ x \ with \ degree \ n \ is \ given \ by:\\ \\ f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{2}x^{2}+a_{1}x+a_{0} \\ \\ where \ n \ is \ a \ nonnegative \ integer \ and \ a_{n}, a_{n-1}, \ldots a_{2}, a_{1}, a_{0} \\ with \ a_{n}\neq 0

Suppose we have the following polynomial function:

f(x)=x^3+x^2-8x-12

If (x-3) is a factor of this polynomial function, we can write f(x) as:

f(x)=\left(ax^2+bx+c\right)\left(x-3\right)

In whose case:

f(3)=0

Evaluating x=3

f(x)=x^3+x^2-8x-12 \\ \\ f(3)=3^3+3^2-8(3)-12 \\ \\ f(3)=27+9-24-12 \\ \\ f(3)=32-32 \\ \\ f(3)=0

So in conclusion:

(x-3) \ is \ a \ factor \ of \ f(x)=x^3+x^2-8x-12

The graph is shown below and is consistent with our conclusion.

<h2>Learn more:</h2>

Complex zeros: brainly.com/question/13728954

#LearWithBrainly

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