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Quadratic function: f(x)= x2+16x+62. Rewrite the function rule in vertex form.

User Hugovdberg
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Final answer:

To rewrite the quadratic function f(x) = x^2 + 16x + 62 in vertex form, we complete the square to get f(x) = (x + 8)^2 - 2, where the vertex of the parabola is (-8, -2).

Step-by-step explanation:

The quadratic function f(x) = x^2 + 16x + 62 can be rewritten in vertex form by completing the square. The vertex form of a quadratic equation is y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. To complete the square, we take the coefficient of x, which is 16, divide it by 2 to get 8, and then square it to get 64. We then add and subtract 64 inside the equation to maintain equality.

Add and subtract 64:
f(x) = x^2 + 16x + 64 + 62 - 64
Now group the perfect square trinomial and the constants:
f(x) = (x^2 + 16x + 64) - 2
Factor the perfect square trinomial:
f(x) = (x + 8)^2 - 2

Now, we have the quadratic function in vertex form, where vertex of the parabola is (-8, -2).

User Morepenguins
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