Question about graphing,vertex,axis of symmetry?

Question

a.) Graph: y=x2-3x+2
b.) Vertex:
c.) Axis of symmetry

Answer 1

it is a U shape open upwardsbecause coefficient of X^2 is positive.

The the lowest point (vertex) is at
x= -b/2a = 3/2 and that is your axis of sym x= 3/2
y = 9/4 - 3/2 + 2 = (9-6+8)/4 =11/4

x intercepts are y=(x-2)(x-1)=0
x= 1 or 2.

Answer 2

You need to put the equation of the parabola in vertex form. You do this by completing the square:

y = x² - 3x + 2

Since the coefficient of the x² term is 1, take half of the coefficient of the x term and square it, then add and subtract that number:

y = x² - 3x + (-3/2)² - (-3/2)² + 2
y = (x² - 3x + 9/4) - 9/4 + 2

Factor the expression in the parentheses and simplify the rest:

y = (x - 3/2)² -1/4

Now you can readily read off the vertex and the axis of symmetry.

Answer 3

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