How do you write # y = 2x^2 + 3x − 5# in vertex form and identify the vertex, y intercept and x intercept?

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Answer 1 · 2018-05-19T18:57:22

#Vertex = (-h, k) = (-3/4, -49/8)#

#"y-int" = -5#

#"x-intercepts" = 1 " and "-5/2#

You must complete the square to solve this:

#y = 2x^2+3x -5#

#y + 5 = 2x^2+3x#

vertex form is:

#y = a(x - h)^2+ k# where #(-h, k)# is the vertex

to complete the square we need a leading coefficient of 1 so factor out the 2

#y + 5 + c = 2(x^2+3/2x + c)#

now #c = (1/2* 3/2)^2 = (3/4)^2 = 9/16#

remember the c value must be multiplied by the 2 we factored out before we add it to the left side!

#y + 5 + (2*9/16) = 2(x^2+3/2x + 9/16)#

#y +49/8 = 2(x+3/4)^2#

finally isolate the y and it is in vertex form:

#y = 2(x+3/4)^2 - 49/8#

#Vertex = (-h, k) = (-3/4, -49/8)#

Y-int is #c# for standard form:

#ax^2+bx+c#

#2x^2+3x -5#

#"y-int" = -5#

you must factor it to find the roots (x-intercepts)

#y = 2x^2+3x -5#

#(x - 1) (2 x + 5)#

#x = 1, -5/2#

So the x-intercepts are #1, -5/2#