How do you graph #f(x)=2x^2-x+1# and identify the x intercepts, vertex?

1 Answer
Jan 21, 2018

See the explanation.

Explanation:

Given:

#color(red)(y=f(x)=2x^2-x+1)#

Our quadratic function is in Standard Form:

#color(blue)( y = f(x) = ax^2+bx+c#, where

#color(green)(a = 2; b=-1 and c=1)#

To find the Vertex, we can use the formula #color(red)([-b/(2a)]#

Hence,

Vertex = #color(red)([-b/(2a)]#

Vertex = #color(red)([-{(-1)/(2*2)}]#

Vertex = #color(red)(1/4 or 0.25}#

This is the x-coordinate value of our Vertex

To find the y-coordinate value of our Vertex,

substitute #color(blue)(x=0.25# in #color(red)(y=f(x)=2x^2-x+1)#

#y = 2*(0.25)^2-(0.25) + 1#

#y = 2*(0.625)-0.25+1#

#y = 0.125-0.25+1#

#y= 0.875#

Hence our Vertex can now be written as an Ordered Pair

Vertex = #color(blue)([0.25, 0.875]#

To find the x-intercepts we set #color(blue)((y=0)# and solve for #color(blue)x#

Hence, we have

#color(red)(y=2x^2-x+1=0)#

We will use the Quadratic Formula to solve for #x#:

#color(green)(x=[-b+- sqrt[b^2 - (4*a*c)]]/(2*a)#

#color(green)(a = 2; b=-1 and c=1)#

Substitute the values in our formula:

#x=[-(-1)+- sqrt[(-1)^2 - (4*2*1)]]/(2*2)#

Simplify to get

#x = 1+-sqrt(-7)/4#

We observe that there are No Real Solutions

Hence, the function does not have x-intercepts.

Additional point of information:

#color(blue)(x=0.25# is known as the Axis of Symmetry

What is the Axis of Symmetry?

The two sides of a graph on either side of the axis of symmetry look like mirror images of each other.

Next analyze the graph below to study the behavior of #f(x)#

enter image source here