Question

How do you find the vertex and intercepts for `y =-3x^2 + 12x – 7`?

Answer 1

`"Vertex "->(x,y)=(+2,5)`

`y_("intercept")=-7`

`x_("intercepts")=2+-sqrt(5/3)" "` Exact values

Approximate values `x_("intercepts")` to 3 decimal places

`x~~+3.291`
`x~~-0.709`

Explanation:

See https://socratic.org/s/aEpUuXgB for a step by step example of method.

Given the standardised form of `y=ax^2+bx+c`

we have the vertex version of the same equation:

`y=a(x+b/(2a))^2+k+c`

Where `k=(-1)xxa(b/(2a))^2`

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

Write as `y=-3(x+12/(-6))^2+[(-3)xx(12/(-6))^2]+7`

`y=-3(x-2)^2+5`

`color(blue)("Vertex "->(x,y)=(+2,5))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine y-intercepts")`

Consider `y=-3x^2+12x-7`

`color(blue)("y-intercept "->y=-7)`

This occurs at `x=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine x-intercepts")`

Note that we have `y=-3x^2+.....`

The coefficient of `x^2` is negative so the graph is of general shape `nn`

The vertex is above the x-axis so as of form `nn` there are x-intercepts

Set `y=0` giving:

`y=-3(x-2)^2+5" "->" "0=-3(x-2)^2+5`

Subtract 5 from both sides

`" "-5=-3(x-2)^2`

Divide both sides by -3

`" "+5/3=(x-2)^2`

Square root both sides

`" "+-sqrt(5/3)=x-2`

add 2 to both sides

`" "x=2+-sqrt(5/3)" "` Exact values

Approximate values to 3 decimal places

`x~~+3.291`
`x~~-0.709`