SOLUTION: Please help. I am pulling my hair out over this? For the equation in exercises 26, create a suitable table of values, and sketch the graph (including the axis of symmetry). Ident

Question 83362: Please help. I am pulling my hair out over this?
For the equation in exercises 26, create a suitable table of values, and sketch the graph (including the axis of symmetry). Identify the axis of symmetry, identify the vertex point, the graph’s direction, and any axis intercepts gleaned from the table or graph.
y=x^2-5x+3

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
for a parabola of the form y=ax^2+bx+c, the equation of the axis of symmetry is x=-b/2a ... in this case x=5/2

the vertex lies on the axis of symmetry and can be found by substituting the x value from the axis of symmetry into the equation

y=(5/2)^2-5(5/2)+3=(25/4)-(25/2)+3=-13/4 ... vertex is ((5/2),(-13/4))

a good table of values would be ... (-1,9), (0,3), (1,-1), (2,-3), (3,-3), (4,-1), (5,3), (6,9)

since the coefficient of the x^2 term is positive, the parabola opens upward

the y-intercept is in the table of values (0,3)

there are 2 x-intercepts, one between (0,3) and (1,-1); and another between (4,-1) and (5,3)
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