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



f(x) = −7(x + 12)2 + 14

I cant figure out the horizontal intercepts.

Verified Answer :

belcatya

f(x) = −7(x + 12)² + 14
 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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