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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.

Sagot :

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=[tex] \frac{-168+- \sqrt{ 168^{2} +4*7*1022} }{14} [/tex]=[tex] \frac{-168+- \sqrt{56840} }{14}[/tex]=[tex] \frac{-168+-14 \sqrt{290} }{14} [/tex]=-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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