Question

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

Answer 1

`"see explanation"`

Explanation:

`"the equation of a parabola in standard form "`

`• y=ax^2+bx+c" has the x-coordinate of the vertex"`

`x_(color(red)"vertex")=-b/(2a)`

`f(x)=x^2-3x+4" is in standard form"`

`"with " a=1,b=-3,c=4`

`rArrx_(color(red)"vertex")=-(-3)/2=3/2`

`"substitute this value into f(x) for y-coordinate"`

`rArry_(color(red)"vertex")=(3/2)^2-3(3/2)+4=7/4`

`rArrcolor(magenta)"vertex "=(3/2,7/4)`

`color(blue)"Intercepts"`

`• " let x = 0 for y-intercept"`

`• " let y = 0 for x-intercepts"`

`x=0toy=4larrcolor(red)" y-intercept"`

`y=0tox^2-3x+4=0`

`"check the value of the "color(blue)"discriminant"`

`Delta=b^2-4ac=(-3)^2-16=-7`

`"since " Delta<0" roots are not real"`

`rArr" there are no x-intercepts"`
graph{x^2-3x+4 [-10, 10, -5, 5]}