Part l: find the x-intercepts of the parabola and write them as ordered pairs. hint: the x-intercepts are the points at which the function crosses the x-axis and y=0. show your work
part ll: write the equation y=(x-4)(x+2) in standard form. show your work.
part lll: with the standard form of the equation from part ll, use the quadratic formula to identify the x-value of the vertex. hint: the x-value of the vertex is -b/2a. show your work
part lv: substitute the x-value of the vertex from part lll into the original equation to find the y-value of the vertex. then, write the coordinates of the vertex

Part 1) The x-intercepts are the points (-2,0) and (4,0)

Part 2)

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

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

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

applying distributive property

where

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

substitute

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

For x=1

substitute and solve for y

therefore

The y-coordinate of the vertex is -9

The coordinate of the vertex is the point (1,-9)

square pyramid

step-by-step explanation:

square prism looks like nothing like that

triangular prism has a rectangular base instead of a square base

tetrahedron has a triangular base