Question

How do you find the vertex and intercepts for `f(x)=3 -8x -4x^2`?

Answer 1

`y_("intercept") = "the constant "=3`

`x_("intercpts") -> x~~0.3229" or "x~~ -2.3229`

`"Vertex "->" "(x,y)=(-1,7)`

Explanation:

Given:`" "y=-4x^2-8x+3`

This equation does not have whole numbers as roots so the formula has to be used to solve for `x_("intercepts")`.

`color(blue)("Determine the y-intercept")`

`color(green)(y_("intercept") = "the constant "=3)`

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`color(blue)("Determine the vertex")`

Write the equation as:
`" "y=-4(x^2 +2x)+3->`part way to vertex form

Consider `2" from "2x`

Apply `(-1/2)xx2 = -1`

`color(blue)(x_("vertex")=-1)`

By substitution:

`color(green)(y_("vertex")=-4(-1)^2-8(-1)+3 = +7)`
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`color(blue)("Determine "x_("intercepts"))`

Using: `y=ax^2+bx+c` where `" "a=-4"; " b=-8"; " c=3`

`" and "x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(+8+-sqrt((-8)^2-4(-4)(3)))/(2(-4))`

`x=(+8+-sqrt(112))/(-8)`

`x=(8+-sqrt(2^2xx28))/-8`

`x=(8+-2sqrt(28))/-8`

`color(green)(x=-1+-sqrt(28)/4 ~~0.3229" or "x~~ -2.3229)`

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