Question

Can anyone someone please help me:((

Answer 1

Answer:

Therefore, the standard for will be:

$y=\frac{1}{4}x^{2}-x-4$

Step-by-step explanation:

The equation of a parabola written as vertical axes is given by:

$(x-h)^{2}=4p(y-k)$ (1)

The vertex of the parabola (h,k) is (2,-5).

The focus (h,k+p) is (2,-4)

Then we can find p knowing that:

h = 2

k = -5

k + p = -4 then p = -4+5 = 1

Putting all these values in equation (1) we will find the equation of the parabola:

$(x-2)^{2}=4(1)(y-(-5))$

$(x-2)^{2}=4(y+5)$ (2)

Now, we need to find the standard form of the equation of the parabola.

Let's recall that standard form is:

$y=ax^{2}+bx+c$  

We just need to work out equation (2) to convert it to the standard form.

$(x-2)^{2}=4(y+5)$

$x^{2}-4x+4=4(y+5)$

$\frac{1}{4}x^{2}-x+1=y+5$

Therefore, the standard for will be:

$y=\frac{1}{4}x^{2}-x-4$

I hope it helps you!