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How do you find the vertex and intercepts for #y=2(x+2)^2-13#?

Answer 1

Find vertex and intercepts for y = 2(x + 2)^2 - 13

y = 2(x + 2)^2 - 13. This is the intercept form, therefor the coordinates of vertex are: Vertex (-2, -13). To find y-intercept, make x = 0 --> #y = 2(2)^2 - 13 = 8 - 13 = -5# To find x-intercepts, make y = 0 --> #0 = 2(x + 2)^2 - 13# #(x + 2)^2 = 13/2 --> x + 2 = +- sqrt13/sqrt2# #x = -2 +- sqrt26/2#

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Answer 2

Eva Adamson

To find the vertex of the quadratic function y = 2(x + 2)^2 - 13, we first need to rewrite the equation in vertex form, which is y = a(x - h)^2 + k, where (h, k) represents the vertex. In this case, a = 2, h = -2, and k = -13. So, the vertex is (-2, -13).

To find the x-intercepts, we set y = 0 and solve for x. So, 2(x + 2)^2 - 13 = 0. Solving this equation gives us the x-intercepts.

To find the y-intercept, we set x = 0 and solve for y. So, y = 2(0 + 2)^2 - 13. Solving this equation gives us the y-intercept.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.