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

1 Answer
Jul 25, 2018

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]}