Answer:
the x-intercepts are
[tex]3\ and\ 4[/tex]
The vertex is the lowest point on the curve
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
if [tex]a>0[/tex] -----> then the parabola open upward (vertex is a minimum)
if [tex]a<0[/tex] -----> then the parabola open downward (vertex is a maximum)
In this problem we have
[tex]y=x^{2} -7x+12[/tex]
[tex]a=1[/tex]
so
the parabola open upward (vertex is a minimum)
Find the x-intercepts of the quadratic equation
Equate the equation to zero
[tex]x^{2} -7x+12=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]x^{2} -7x=-12[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]x^{2} -7x+12.25=-12+12.25[/tex]
[tex]x^{2} -7x+12.25=0.25[/tex]
Rewrite as perfect squares
[tex](x-3.5)^{2}=0.25[/tex]
square root both sides
[tex](x-3.5)=(+/-)0.5[/tex]
[tex]x=3.5(+/-)0.5[/tex]
[tex]x=3.5+0.5=4[/tex]
[tex]x=3.5-0.5=3[/tex]
therefore
the x-intercepts are
[tex]3\ and\ 4[/tex]
