How do you find the vertex and the intercepts for #y= -3(x-1)(x+5)#?

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Answer 1 · 2016-03-29T00:38:52

Vertex is at #(-2, 27)#
Intercepts at #(1, 0)#, #(-5, 0)#,#(0, 15)#

From the given #y=-3(x-1)(x+5)#
Expand the right side so that it will take the form #y=ax^2+bx+c#

#y=-3(x-1)(x+5)#

#y=-3(x^2+4x-5)#

#y=-3x^2-12x+15#

Now it takes the form #y=ax^2+bx+c#

with #a=-3# and #b=-12# and #c=15#

We can now solve for the vertex #(h, k)# using the formula

#h=-b/(2a)# and #k=c-b^2/(4a)#

#h=(-(-12))/(2(-3))=12/-6=-2#

#k=c-b^2/(4a)=15-(-12)^2/(4(-3))=15+144/12=15+12=27#

Vertex #(h, k)=(-2, 27)#

To solve for the intercept, let us use the given #y=-3(x-1)(x+5)#

Solve for x-intercept by setting #y=0#

#y=-3(x-1)(x+5)#

#0=-3(x-1)(x+5)#

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

Equate both factors to 0
First factor
#(x-1)=0#
#x=1# there is an intercept at #(1, 0)#

Second factor
#x+5=0#
#x=-5# there is an intercept at #(-5, 0)#

Use again the original equation #y=-3(x-1)(x+5)# to solve for y-intercept and setting #x=0#

#y=-3(x-1)(x+5)#

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

#y=-3(-1)(5)#

#y=15#

there is an intercept at #(0, 15)#

The graph of #y=-3(x-1)(x+5)# where you can see the
vertex #-2, 27)#, and intercepts at #(1, 0)#, #(-5, 0)#,#(0, 15)#

graph{y=-3(x-1)(x+5)[-60,60,-30,30]}

God bless .... I hope the explanation is useful.