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]}
