How do you find the vertex and intercepts for y=(x+5)(x+3)y=(x+5)(x+3)?

1 Answer
Dylan D.
Apr 21, 2018

the xx-intercepts are -5−5 and -3−3, the y-intercept is 1515 and the vertex is (-4,-1)(−4,−1)

Explanation:

Because the equation is in the form y=a(x-p)(x-q)y=a(x−p)(x−q), we know the xx-intercepts are pp and qq
Therefore, the xx-intercepts are at -5−5 and -3−3
Note the negative signs and the fact that in the case a=1a=1 so it is committed.
This makes sense because the xx-intercepts are whenever y=0y=0. This occurs when either bracket is 00.

The xx-value of the vertex is the midvalue of the intercepts.
let V_xVx be the xx-value of the vertex
V_x=((-5)+(-3))/2Vx=(−5)+(−3)2
V_x=-4Vx=−4
Subbing in x=-4x=−4 to solve for the yy-value of the vertex:
Let this be V_yVy:
V_y=((-4)+5)((-4)+3)Vy=((−4)+5)((−4)+3)
V_y =(1)(-1)Vy=(1)(−1)
V_y = -1Vy=−1

Therefore, the vertex is (-4,-1)(−4,−1)

To find the yy-intercept we substitute x=0x=0 into the equation.
y=(5)(3)y=(5)(3)
y=15y=15

In conclusion, the xx-intercepts are -5−5 and -3−3, the yy-intercept is 1515 and the vertex is (-4,-1)(−4,−1)

Included below are graphs of the equation:
graph{(x+5)(x+3) [-7.757, 3.6, -2.454, 3.224]}
graph{(x+5)(x+3) [-49.16, 45, -23.16, 23.9]}

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