# How do you find the vertex and intercepts for #y=3x^2+5x+8#?

##### 1 Answer

Jul 4, 2018

#### 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=3(x^2+5/3x+8/3)#

#color(white)(y)=3(x^2+2(5/6)x color(red)(+25/36)color(red)(-25/36)+8/3)#

#color(white)(y)=3(x+5/6)^2+3(-25/36+96/36)#

#color(white)(y)=3(x+5/6)^2+71/12larrcolor(blue)"in vertex form"#

#rArrcolor(magenta)"vertex "=(-5/6,71/12)#

#"to find the x-intercepts let y = 0"#

#3(x+5/6)^2+71/12=0#

#3(x+5/6)^2=-71/12#

#"this has no real solutions hence no x-intercepts"#

#"for y-intercept let x = 0"#

#y=0+0+8=8larrcolor(red)"y-intercept"#

graph{3x^2+5x+8 [-20, 20, -10, 10]}