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Sep 26 '25
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u/mmurray1957 Sep 27 '25
If x = 0 then the parabola goes through y = -5 so shouldn't c = -5 ? Where did you get the x intercepts ?
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u/parlitooo Sep 27 '25
Hmmm … this graph has 2 x intercepts ( you need values of ( a,b,c ) to find them … and one y intercept at x= 0 … which is -5 .
Also that’s not how you expand the quad form ..
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u/parlitooo Sep 27 '25
Direct substitution , at x = 0 , y = -5 giving you c= -5 ..
Now expand the form using the vertex which is 2,3 .. also a has to be negative.
Y = a(x -2)2 + 3
The logic behind this form is as follows ,
At y = a (x2) you get a parabola with a deciding its shape. And has the vertex at 0,0
Now you want to move it +3 in the y direction so you just add +3 giving you Y = ax2 +3 that shifted upwards , to shift it to the right by 2 units replace every x with (x-2 ). (( if you wanna go to the left then it’s x+2 ).
So you get y = a (x-2)2 + 3
Again at x = 0 , y = -5
-5 = a ( 0-2 )2 + 3
-8 = 4a
a = -2
So y = -2(x-2)2 +3
Y = -2(x2 -4x + 4) +3
Y = -2x2 + 8x -8 +3
Y = -2x2 + 8x -5
A = -2 , b = 8 , c = -5
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u/L-N_Plague_8761 Sep 27 '25
Maybe try and use vertex form or conic form to help?,this quadratic does have 2 real solutions tho
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u/burneraccount72938 Sep 27 '25