How do you use the important points to sketch the graph of #y=3x^2+5x-2#?

1 Answer
Sep 5, 2017

The important points for sketching a quadratic are where it crosses or turns at the #x# axis, crosses the #y# axis and the vertex of the parabola.

It crosses the #x# axis at the roots:

#3x^2 + 5x - 2 = 0#

Factoring:

#( 3x -1 )( x + 2) = 0 => x = 1/3 and x = -2#

It crosses the #y# axis where #x = 0#:

#3(0)^2 + 5(0) -2 = 0 => y = -2#

Vertex is at# ( -5/6 , -47/36 )#

To find vertex of #3x^2 + 5x -2 #:

Place a bracket around the terms containing the variable.

(#3x^2 + 5x)- 2# factor out the coefficient of #x^2# if this is not# 1#
#3(x^2 + 5/3x) - 2#
Add the square of half the coefficient of #x# inside the bracket and subtract it outside the bracket.
#3(x^2 + 5/3x + (5/6)^2 ) - (5/6)^2 - 2#

Arrange
#(x^2 + 5/3x + (5/6)^2 )# into the square of a binomial and simplify# - (5/6)^2 - 2#:

#3( x + 5/6 )^2 - 47/36#

This is now in the form #a( x - h )^2 + k#

Where #h# is the axis of symmetry and #k# is the maximum or minimum value of the function.

So vertex is #( -5/6 , -47/36 )#

Points for drawing graph are:

#( 1/3 , 0 ) ( -2 , 0 )# roots of equation.

#( 0 , -2 )# #y# axis intercept.

#(- 5/6 , - 47/36 )# vertex