SOLUTION: Please help me. I have no clue on what numbers to use for this parabola since my Y intercepts come out as big numbers. I need to sketch the parabola with the equation 4x^2-16x+15

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Question 897132: Please help me. I have no clue on what numbers to use for this parabola since my Y intercepts come out as big numbers.
I need to sketch the parabola with the equation 4x^2-16x+15 and label the vertex, axis of symmetry, x and y intercepts. Should I divide by 4?

Found 3 solutions by rothauserc, MathTherapy, richwmiller:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website!
graph is

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!
Please help me. I have no clue on what numbers to use for this parabola since my Y intercepts come out as big numbers.
I need to sketch the parabola with the equation 4x^2-16x+15 and label the vertex, axis of symmetry, x and y intercepts. Should I divide by 4?

Standard form of a parabolic equation:
The x-coordinate of the vertex occurs at: . This is also the axis of symmetry.
The y-coordinate of the vertex occurs at: , where x is the value of the x-coordinate of the vertex
The x intercepts, or roots occur at the solutions to:
The y-intercept occurs at . Looking at the equation, it should be:
**FYI
If you intend to create the vertex form of the equation, then you'd need to divide through by 4.
Otherwise, that's too much work.

Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website!
y=4x^2-16x+15
y = (2x-5)*(2x-3)
let x=0 y=15
let x=1 y=3
let x=2 y=-1
min{y = 4 x^2-16 x+15} = -1 at x = 2
parabola
focus | (2, -15/16)~~(2, -0.9375)
vertex | (2, -1)
semi-axis length | 1/16 = 0.0625
focal parameter | 1/8 = 0.125
eccentricity | 1
directrix | y = -17/16