Question 172897
{{{y= (2x-7) (2x-1) }}}
Expand:
{{{y=4x^2-14x-2x+7}}}
{{{y=4x^2-16x+7}}}
Transpose constant to the left, likewise "copy" the "a" constant of a quadratic eqn --->{{{highlight(a)x^2+bx+c}}}to the left. factor out the right term:
{{{y-7+4*highlight(4)=4(x^2-4x)}}}
Complete the square by taking "half" of '4' on the right term,in this ---> {{{4/2=2}}}, then squared it---->{{{2^2=4}}}.
{{{y-7+16=4(x^2-4x+highlight(4))}}}
{{{y+9=4(x-2)^2}}}
{{{y=4(x-2)^2-9}}} --> it follows vertex form, {{{a(x-h)^2+k}}}
where--->{{{system(h=x,k=y)}}}
So, therefore, the vertex is at (2,-9).
Axis of symmetry is the x-intercept of the vertex: --->{{{highlight(x=2)}}}
For Y-Intercept:
{{{F(x)=0}}}
{{{y=4(0)^2-14(0)+7}}}
{{{highlight(y=7)}}}
For the X-Intercepts: We solve the eqn:
{{{0=4(x-2)^2-9}}}
{{{9=4(x-2)^2}}}------>{{{9/4=cross(4)(x-2)^2/cross(4)}}}
{{{+-sqrt(9/4)=x-2}}}
{{{x=2+-1.5}}} ---->{{{system(x=2+1.5=highlight(3.5),x=2-1.5=highlight(0.50))}}}
As you see in the graph:
{{{drawing(400,400,-5,5,-15,15,grid(1),graph(400,400,-5,5,-15,15,4x^2-16x+7),circle(3.5,0,.12),circle(0.5,0,0.12),blue(circle(2,-9,.12)),green(circle(0,7,.12)))}}} ---->Vertex (2,-9)
Thank you,
Jojo