y=-x^2+4x+5

Question

I need to find the y-intercept and x- intercept and the vertex

Answer 1

First, for the x-intercept, we just set x = 0:

 

y = -0² + 4(0) + 5 = 5 ---> y = 5

 

For the y-intercept, we will factor it and set y = 0:

 

0 = -x² +4x + 5

0 = (x + 1)(-x + 5)

x = -1, 5

 

To find the vertex, it is easiest to complete the square:

 

-x² +4x + 5 = -[x² - 4x -5] = -[x² -4x +4 - 4 - 5] = -[(x - 2)² - 4 - 5] = -(x - 2)² + 9

 

Therefore the vertex occurs at x = 2, since (x - 2) is in the squared term.  Now we can plug x = 2 to find the y-coordinate of the vertext

 

y = -(2)² + 4(2) + 5 = 9 ---->  Vertex at (2,9)

 

 

J.T.