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

##### 1 Answer

Jun 29, 2017

#### 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]}