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How do you find the vertex and intercepts for #y = (1/8)(x – 5)^2 - 3#?

Answer 1

Savannah Anderson

the vertex is at #(5,-3)#

Your quadratic equation is of the form #y=a (x-h)^2+k# The vertex is at the point #(h,k)#

#y= 1/8 (x-5)^2 + (-3)#

the vertex is at #(5,-3)# graph{1/8(x-5)^2-3 [-2.03, 14.67, -6.53, 1.814]}

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Answer 2

Abigail Allen

To find the vertex of the parabola represented by the equation (y = \frac{1}{8}(x - 5)^2 - 3), we use the formula for the vertex of a parabola in vertex form, (y = a(x - h)^2 + k), where the vertex is given by the point ((h, k)). In this equation, (h) and (k) are the coordinates of the vertex. By comparing the given equation to the vertex form, we find that (h = 5) and (k = -3). Therefore, the vertex is at the point ((5, -3)).

To find the x-intercepts, we set (y = 0) and solve for (x). This gives us: [0 = \frac{1}{8}(x - 5)^2 - 3] [3 = \frac{1}{8}(x - 5)^2] [24 = (x - 5)^2] [±√24 = x - 5] [x = 5 ±√24]

So, the x-intercepts are (x = 5 + √24) and (x = 5 - √24).

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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.