How do you graph #Y=X^2-4X-5#?

1 Answer
Sep 17, 2015

This is a quadratic function of form #y=ax^2+bx+c#, with a=1, b=-4 and c=-5.
The graph will be a parabola.

Since a is positive, the arms go up.
To find the y-intercept, let x = 0 and find y = c = -5.
To find the x-intercepts, let y = 0 and use either factorizing or the quadratic formula to find the roots x = 5 or x = -1.

To find the turning point, set the derivative to zero.
#therefore2x-4=0 =>x=2>#
Then find y(2) = -9.

Then put all of this together to draw the graph as shown below :

graph{x^2-4x-5 [-22.49, 22.5, -11.24, 11.26]}