Question

How do you find the vertex and the intercepts for `y=-4x^2+8x+13`?

Answer 1

`color(green)(y_("intercept")->(x,y)=(0,13))`

`color(blue)(x_("intercept")~~ 3.062" and "~~-1.062 larr" to 3 decimal places")`
`color(blue)(x_("intercept") =1+-sqrt(17)/2 larr" exact answer")`

`color(magenta)("Vertex"->(x,y)=(1,17))`

Explanation:

Given:`" "y=-4x^2+8x+13`

`color(blue)("Determine the general shape of the curve")`

The coefficient of `x^2` is negative. Consequently the graph is pf form `nn`. Thus the vertex is a maximum

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`color(blue)("Determining y-intercept")`

Reading directly off the equation the y-intercept is +13
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`color(blue)("Determining the vertex")`

Write as `-4(x^2-8/4x)+13`

`x_("vertex")=(-1/2)xx(-8/4) = +1`

By substitution: `" "y_("vertex")=-4(1)^2+8(1)+13 = 17`

Vertex`->(x,y)=(1,17)`

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`color(blue)("Determining x-intercept")`

The vertex is above the x-axis so as the general form is `nn` x-intercepts exist.

Compare to the standardised form `y=ax^2bx+c`

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

`a=-4"; "b=8"; "c=13` giving:

`x=(-8+-sqrt(8^2-4(-4)(13)))/(2(-4))`

`x=(-8+-sqrt(272))/(-8)`

`x=1+-(sqrt(2^2xx2^2xx17))/(-8)`

`x=1+-(4sqrt(17))/-8" " ->" " 1+-sqrt(17)/2 larr" exact answer"`

You may write `4/(-8)` as `1/2` as its sign does not matter due to the `+-` preceding that part of the equation.

`x~~ 3.062" "x~~-1.062`