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]}