Question

How do you find the vertex and the intercepts for `y= x^2-8x+5`?

Answer 1

`"vertex "=(4,-11)`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form "color(blue)"complete the square"`

`y=x^2+2(-4)x+16-16+5`

`color(white)(y)=(x-4)^2-11larrcolor(blue)"in vertex form"`

`color(magenta)"vertex "=(4,-11)`

`"to obtain the y-intercept let x = 0"`

`y=0-0+5=5larrcolor(red)"y-intercept"`

`"to obtain the x-intercepts let y = 0"`

`(x-4)^2-11=0`

`(x-4)^2=11`

`color(blue)"take the square root of both sides"`

`x-4=+-sqrt11larrcolor(blue)"note plus or minus"`

`"add 4 to both sides"`

`x=4+-sqrt11larrcolor(red)"exact solutions"`

`x~~0.68" or "x~~7.32" to 2 dec. places"`