Question

How do you find the vertex and intercepts for `f(x)=4x^2-32x+63`?

Answer 1

I have given the answer to and shown the method to obtain
`color(blue)(x_("vertex")= +4)`
I have given the method to find the rest.

Explanation:

`color(blue)("To find " x_("vertex"))`

Given: `y=4x^2-32x+63 ........(1)`

Write as: `4(x^2-32/4 x)+63`

Consider the `-32/4 " from " -32/4x`

`x_("vertex")=(-1/2)(-32/4) = +32/8 =4...(2)`
This compares to the graph
.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Method from this point:")`

`color(blue)("To find " y_("vertex"))`

Substitute (2) into (1) to solve for `y_("vertex")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To find " y_("intercept"))`

`y_("intercept") = 63` this is the constant at the end of equation (1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("To find " x_("intercept"))`

Substitute `y=0` in equation (1) and solve for `x`

If you are not sure about factoring use the formula

Standard form: `y=ax^2+bx+c`

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

You can see from the graph that your answers should be close to `3 1/2 " and " 4 1/2 color(red)(" "underline("These are estimates"))`