Answer:
Part 1) The x-intercepts are the points (-2,0) and (4,0)
Part 2) [tex]y=x^{2} -2x-8[/tex]
Part 3) The x-coordinate of the vertex is 1
Part 4) The y-coordinate of the vertex is -9 and the coordinate of the vertex is the point (1,-9)
Step-by-step explanation:
we have
[tex]y=(x-4)(x+2)[/tex]
Part 1) Find the x-intercepts of the parabola and write them as ordered pairs
The x-intercepts are the values of x when the value of y is equal to zero
so
For y=0
[tex](x-4)(x+2)=0[/tex]
For x=4 and x=-2 the equation is equal to zero
therefore
The x-intercepts are the points (-2,0) and (4,0)
Part 2) Write the equation y=(x-4)(x+2) in standard form
The quadratic equation in standard form is equal to
[tex]y=ax^{2} +bx+c[/tex]
applying distributive property
[tex]y=(x-4)(x+2)\\\\y=x^{2}+2x-4x-8\\\\ y=x^{2} -2x-8[/tex]
where
[tex]a=1, b=-2,c=-8[/tex]
Part 3) With the standard form of the equation from part ll, use the quadratic formula to identify the x-value of the vertex
we know that
the x-value of the vertex is -b/2a
we have
[tex]a=1, b=-2,c=-8[/tex]
substitute
[tex]-\frac{b}{2a}=-\frac{(-2)}{2(1)}=1[/tex]
therefore
The x-coordinate of the vertex is 1
Part 4) Substitute the x-value of the vertex from part lll into the original equation to find the y-value of the vertex.
we have
[tex]y=(x-4)(x+2)[/tex]
For x=1
substitute and solve for y
[tex]y=(1-4)(1+2)[/tex]
[tex]y=(-3)(3)[/tex]
[tex]y=-9[/tex]
therefore
The y-coordinate of the vertex is -9
The coordinate of the vertex is the point (1,-9)
