How do you graph # y=-(x+3)^2 + 5#?

1 Answer
Meave60
Dec 17, 2017

Vertex: #(-3,5)#

Y-intercept: #(0,-4)#

X-intercepts: #(-3+sqrt5,0),##(-3-sqrt5,0)#

Approximate x-intercepts: #x~~(-0.7639,0),##"and"##~~(-5.236,0)#

Explanation:

Graph:

#y=-(x+3)^2+5# is a quadratic equation in vertex form:

#y=a(x-h)^2+k#,

where:

#a=-1#, #h=-3#, and #k=5#

The vertex, #(h,k)#, is #(-3,5)#

To find the y-intercept, substitute #0# for #x# and solve for #y#.

#y=-1(0+3)^2+5#

#y=-9+5#

#y=-4#

Y-intercept: #(0,-4)#

To find the x-intercepts, substitute #0# for #y# and solve for #x#.

#-(x+3)^2+5=0#

#-(x+3)^2=-5#

Divide both sides by #-1#.

#(x+3)^2=(-5)/(-1)#

Simplify.

#(x+3)^2=5#

Take the square root of both sides.

#x+3=+-sqrt5#

Subtract #3# from both sides.

#x=-3+-sqrt5#

X-intercepts: #(-3+sqrt5,0),##(-3-sqrt5,0)#

#x~~(-0.7639,0),##"and"##~~(-5.236,0)#

Summary:

Vertex: #(-3,5)#

Y-intercept: #(0,-4)#

X-intercepts: #(-3+sqrt5,0),##(-3-sqrt5,0)#

Approximate x-intercepts: #x~~(-0.7639,0),##"and"##~~(-5.236,0)#

Plot the points and sketch a parabola through the points. Do not connect the dots.

graph{y=-(x+3)^2+5 [-10, 10, -5, 5]}

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