Question

How do you use the important points to sketch the graph of `y = 4x^2 + 5x - 1`?

Answer 1

Below

Explanation:

To graph a parabola, it is a good idea to find:

• vertex
• x-intercepts
• y-intercepts

X-intercepts
When `y=0`,
`x=(-5+-sqrt(25-4(4)(-1)))/(2times4)`
`x=(-5+-sqrt41)/8`
`x=(-5+sqrt41)/8` and `x=(-5-sqrt41)/8`

Y-intercepts
When `x=0`, `y=-1`

Vertex
`x=-b/(2a)` where the quadratic equation is in the form `ax^2+bx+c`
`x=(-5)/(2times4)`
`x=-5/8`
Sub `x=-5/8` into `y=4x^2+5x-1` to get the y-coordinate
`y=-2 9/16`
The vertex is `(-5/8,-2 9/16)`

Plotting your vertex and intercepts, you should get the graph below

graph{4x^2+5x-1 [-10, 10, -5, 5]}