How do you write the quadratic in vertex form given vertex is (-2,6) and y intercpet is 12 ?

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How do you write the quadratic in vertex form given vertex is (-2,6) and y intercpet is 12 ?

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Answer 1 · 2015-05-30T22:22:46

Quadrati equation $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ .

c = 12, find a and b.

x of vertex: $(- \frac{b}{2} a) = - 2 \to b = 4 a$ $(-b/2a) = -2 -> b = 4a$

y of vertex: #f(-2) = 6 = 4a - 2b + 12 --> 4a - 8a + 12 = 6 ->

-> $a = \frac{6}{4} = \frac{3}{2}$ $a = 6/4 = 3/2$

Equation $y = \frac{3}{2} (x^{2}) + 6 x + 12$ $y = 3/2(x^2) + 6x + 12$

Check:

x of vertex: $(- \frac{b}{2 a}) = - \frac{6}{3} = - 2$ $(-b/(2a)) = -6/3 = -2$ OK
$f (- 2) = (\frac{3}{2}) 4 - 12 + 12 = 6$ $f(-2) = (3/2)4 - 12 + 12 = 6$ OK

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How do you write the quadratic in vertex form given vertex is (-2,6) and y intercpet is 12 ?

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