Question

how do you put "y-5=-2(x+1)^2" into intercept form? and "y=(x+2)(x-3)" into vertex form?

Answer 1

i put "y-5=-2(x+1)^2" into standard form and got y=-2x^2-4x+3

Answer 2

I got y=x^2-x-6 for the other one

Answer 3

For the first one, the vertex form "y-k=a(x-h)" where h,k is the vertex.

I get y=(x+2)(x-3) = x^2-x-6
So we need to get an (x-h)^2 term, so

y+6 = x^2 + x complete the square

y+6+(1/2)^2 = x^2 +x + (1/2)^2
y+25/4 = (x+1/2)^2 is the vertex form, so

h is -1/2, k is 25/4 , the vertex is at point (-1/2, 25/4).


For the intercept form one, I don't really know how to get it into that form.

I don't even remember that form being taught when I was doing grade 12 math. You could find the zeros (intercepts) maybe through synthetic division? I tried the possible factors and none seemed to work out. I'll take a crack at it again and see if I missed anything. Stay tuned.

Answer 4

Sorry, the first line above should read "For the first one, the vertex form "y-k=a(x-h)^2" where h,k is the vertex". Just forgot to add the exponent.

Answer 5

Hi Chowbelly, I tried working with it again, the only way I could find the intercepts was through completing the square again.

Start out with y=-2x^2+4x+3, find zeros

0= -2x^2 + 4x +3
-3 = -2x^2 +4x
-3 = -2(x^2 -2x)
-3-2 = -2(x^2 -2x +1) this is the "complete the square" step

-5 = -2(x - 1)^2
5/2 = (x-1)^2
take square roots of both sides we
get
+/-(5/2)^1/2 = x-1,

x=1+(5/2)^1/2, 1-(5/2)^1/2

so the intercept form would be y = (x-(1+(5/2)^1/2))(x-(1-(5/2)^1/2))

or y= (x + .5811)(x - 2.5811)

Hope this helps.

Answer 6

thank you!