How do you find the intercept and vertex of #y = (x − 3)(4x + 2)#?

#color(blue)("Determine the x-intercepts")#

x-intercept is at #y=0#

Set #y=0=(x-3)(4x+2)#

Consider the case #(x-3)=0 => x=3#
Consider the case #(4x+2)=0->2(2x+1)=0=>x=-1/2#
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#color(blue)("Determine the vertex")#

If you multiply out the brackets you have #+4x^2 ....#. This means that the graph is of form #uu# thus the vertex is a minimum.

#x_("vertex")# will be half way between the x-intecpts

#x_("vertex")=(3+(-1/2))/2 = 2.5/2=1.25 -> 5/4#

Thus #y_("vertex") = (5/4-3)(5+2) = -7/4xx7=-49/4#
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#color(blue)("Determine the y-intercept")#

Multiplying out the constants in the brackets we have

#(-3)xx(+2) = -6#

Thus #y_("intercept")=-6#

Notice that the section for this question is 'Vertex Form ....'

#y=(x-3)(4x+2)=4x^2-10x-6#

Set #y=4(x^2-10/4x)-6#

Halve the coefficient of #x#

#y=4(x^2-10/8x)-6#

Remove the #x# from #-10/8x# and move the squaring to outside the brackets

#y=4(x-10/8)^2-6# This will not produce the original equation #color(white)("ddddddddddddddddd")#as we have not includes a correction
#color(white)("dddddddddddddddddd")#process.

#y=color(red)(4)(xcolor(green)(-10/8))^2+k-6# Now it will!

Note that #color(red)(4)xxcolor(green)((-10/8)^2)+k=0#

#=> k=-25/4#

So #k-6->-25/4-24/4=-49/4#

So the final format is:

#y=4(x-5/4)^2-49/4#