Your answer should be in vertex form.

rompentskj
2021-12-03
Answered

Determine the equation of the quadrac function with vertex (2,4) and passing through the point (-1, -14).

Your answer should be in vertex form.

$f\left(x\right)=$

Your answer should be in vertex form.

You can still ask an expert for help

Coon2000

Answered 2021-12-04
Author has **15** answers

Let the vertex form of quadratic function

be

the vertex.

We are given: vertex is (2, 4)

We are aslo given that function

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Let $A\left(\alpha \right)=(\alpha I+(1-\alpha )B)-1$ be the inverse of a convex combination of the n-by-n identity matrix I and a positive semidefinite symmetric matrix B, where the combination is given by $\alpha \in (0,1]$ . The matrix B is not invertible. For some n-dimensional vector c, I want to find $\alpha$ such that

${c}^{T}A\left(\alpha \right)(I-B)A\left(\alpha \right)c=0$

Is there a closed-form solution for$\alpha$ ?

Is there a closed-form solution for

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Factor the following quadratics.

a)$2{x}^{2}+7x+6$

b)$4{x}^{2}-13x+9$

c)$3{x}^{2}-10x+8$

d)$2{x}^{2}-3x-2$

e)$3{x}^{2}+17x+20$

f)$4{x}^{2}-25x+25$

a)

b)

c)

d)

e)

f)

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Range of p satisfying quadratic inequality

What is the set of values of p for which$p({x}^{2}+2)<2{x}^{2}+6x+1$ , for all real values of x?

What is the set of values of p for which

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A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function.

-The quadratic coefficients is

-The linear coefficients is

-The constant term is

-The quadratic coefficients is

-The linear coefficients is

-The constant term is

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Solve $\sqrt{{x}^{2}+8x+7}+\sqrt{{x}^{2}+3x+2}=\sqrt{6{x}^{2}+19x+13}$

asked 2021-08-17

how do i solve?