Determine the coordinates of the vertex and horizontal intercepts of the parabola. (If an answer does not exist, enter DNE.)

1 answer:

7 0

<span>f(x) = −7(x + 12)² + 14</span>
the coordinates of the vertex are (-12;14)
f(x)=-7(x²+24x-144)+14
f(x)=-7x²-168x+1008+14
f(x)=-7x²-168x+1022=0
7x²+168x-1022=0
x= $\frac{-168+- \sqrt{ 168^{2} +4*7*1022} }{14}$ = $\frac{-168+- \sqrt{56840} }{14}$ = $\frac{-168+-14 \sqrt{290} }{14}$ =-12+-√290
x₁=-12-√290
x₂=-12+√290
Answer: the coordinates of the vertex are (-12;14),
horizontal intercepts are x₁=-12-√290, x₂=-12+√290

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