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How do you find the axis of symmetry, vertex and x intercepts for #y=x^2-6x+8#?

Answer 1

Vertex (3, -1)
x-intercepts: 2 and 4

#y = x^2 - 6x + 8# x-coordinate of axis of symmetry, and of vertex: #x = - b/(2a) = 6/2 = 3# y-coordinate of vertex: #y(3) = 9 - 6(3) + 8 = -9 + 8 = - 1# Vertex (3, -1) To find the 2 x-intercepts, make y = 0 and solve: #y = x^2 - 6x + 8 = 0# Find 2 real roots (x- intercepts) knowing the sum (-b = 6) and the product (c = 8). They are 2 and 4.

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Answer 2

Hazel Anglin

To find the axis of symmetry, use the formula x = -b / (2a), where a = 1, and b = -6. Substituting these values, x = -(-6) / (2*1) = 3.

To find the vertex, plug the x-coordinate of the axis of symmetry into the quadratic equation to find the y-coordinate. So, when x = 3, y = (3)^2 - 6(3) + 8 = 1.

To find the x-intercepts, set y = 0 in the quadratic equation and solve for x. So, 0 = x^2 - 6x + 8 can be factored as (x - 4)(x - 2) = 0. Thus, x = 4 and x = 2 are the x-intercepts.

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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.