Can someone please help me write the standard form of the quadratic function that has the indicated vertex and whose graph passes through the given point.

Vertex: (-5/2,0)
Point: (-7/2,-16/3)

Question

Can someone please help me write the standard form of the quadratic function that has the indicated vertex and whose graph passes through the given point. 

Vertex: (-5/2,0)
Point: (-7/2,-16/3)

campbell_st · Answer

well the standard vertex form is 

$$y = a(x -h)^2 + k$$

where the vertex is (h, k)

so you can substitute the vertex directly into the equation. 

next to find a, 

after substituting the vertex, substitute the point into the equation to get the value of a. 

lastly, to get the equation in standard form 

with the vertex substituting and having the value of a, 

expand the perfect square and distribute a

then collect like terms.. 

hope this helps

anonymous · Answer

would just substituting the vertex be y=a(x-4)^2-1? @campbell_st

anonymous · Answer

how do i substitute the point if there are two numbers for the point?

campbell_st · Answer

nope the vertex is (-5/2, 0) so 

h = -5/2 and k = 0

campbell_st · Answer

and if you have a point (x, y) 

and an equation y =a(x -h)^2 + k

you have an x and y value in the equation, replace then with the values from the point...

anonymous · Answer

sorry lol i was looking at a different problem so now i have -16/3=a(7/2-(-5/2)^2+0 is that correct? @campbell_st

campbell_st · Answer

thats correct, so solve for a

then you'll have your equation

anonymous · Answer

what do i need to do to solve for a??

campbell_st · Answer

well simplify 

-16/3 = a(7/2 - (-5/2))^2

or

-16/3 = a(72/ + 5/2)^2

-16/3 = a(12/2)^2

etc...

anonymous · Answer

so then a would be -4/27?

anonymous · Answer

so the standard form with the given information would be -16/3=-4/27(7/2-(-5/2)^2?

anonymous · Answer

@campbell_st

anonymous · Answer

this is what I have, how would I write this in the proper standard form?

campbell_st · Answer

ok..so you have the value of a then 

$$y = -\frac{4}{27}(x - \frac{5}{2})^2$$

is the vertex form... 

distribute to get the standard form 

so its

$$y =- \frac{4}{27}(x^2 - 5x + \frac{25}{4})$$

just distribute for the final standard form...

anonymous · Answer

so would the final standard form be y=-(2x-5)^2/27? i dont know if i did this right

campbell_st · Answer

well in standard form, I'd expect you'd need to distribute the -4/27

anonymous · Answer

when i try distrbuting the -4/27 i keep getting -(2x-5)^2/27

campbell_st · Answer

well my starting point is 

$$y = -\frac{4}{27}(x^2 - 5x + \frac{25}{4}) = \frac{-4x^2 + 20x -25}{27}$$

just as a guess...

anonymous · Answer

well then wouldnt it be -4x^2+20x-100/27?