Write an equation for a quadratic function in vertex

Ответ:

dri16947

29.07.2021

$\displaystyle f(x)=-\frac{1}{2}(x+1)^2+6$

Step-by-step explanation:

We want to write the equation for a quadratic function in vertex form with vertex at (-1, 6) that passes through the point (-3, 4).

Vertex form is given by:

$\displaystyle f(x)=a(x-h)^2+k$

Where (h, k) is the vertex and a is the leading coefficient.

Since our vertex is at (-1, 6), h = -1 and k = 6. Substitute:

f(x)=a(x-(-1))^2+(6)

Simplify:

f(x)=a(x+1)^2+6

Next, since the quadratic passes through the point (-3, 4), f(x) = 4 when x = -3. Substitute:

(4)=a(-3+1)^2+6

Solve for a. Simplify:

-2=a(-2)^2

Hence:

$\displaystyle -2=4a\Rightarrow a=-\frac{1}{2}$

Therefore, our function in vertex form is:

$\displaystyle f(x)=-\frac{1}{2}(x+1)^2+6$

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ayyyyyyyydubs

07.02.2023

tuesday and thursday

But the other 2 idk

I'll g o o g l e

Step-by-step explanation:

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