Answer:
If the equation is in the form
y+=+ax%5E2+%2B+bx+%2B+c , then the x component
of the vertex is at
and, plugging the point into general equation,
y+=+ax%5E2+%2B+bx+%2B+c
-5+=+a%2A6%5E2+%2B+b%2A6+%2B+c
(2) -5+=+36a+%2B+6b+%2B+c
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The y-intercept is at (0,175), so plugging this into
general equation:
175+=+a%2A0%5E2+%2B+b%2A0+%2B+c
(3) 175+=+c
Now I have 3 equations and 3 unknowns, so it's solvable
From (1),
(1) 6+=+-%28b%2F2a%29
b+=+-12a
Plug this into (2)
(2) -5+=+36a+%2B+6b+%2B+c
(2) -5+=+36a+%2B+6%2A%28-12a%29+%2B+c
also, from(3),
(2) -5+=+36a+%2B+6%2A%28-12a%29+%2B+175
36a+-+72a+=+-175+-5
-36a+=+-180
a+=+5
and, since
b+=+-12a
b+=+-12%2A5
b+=+-60
Now I can write the actual equation
y+=+5x%5E2+-+60x+%2B+175
Set y+=+0 to find x-intercepts
Then divide through by 5
x%5E2+-+12x+%2B+35+=+0
I can see right away that it factors into
%28x+-+5%29%28x+-+7%29+=+0
x+=+5 and x+=+7 are the solutions
So, the x-intercepts are at (5,0) and (7,0)
Step-by-step explanation:
