How do you graph the parabola #f(x)=-3x^2+x-5# using vertex, intercepts and additional points?

1 Answer
Jan 23, 2018

Answer:

The #f(x)# intercept is at #f(x)=-5#, and the #x# intercept is at {TBA}. The vertex is at the point #(1/6, 4 3/4)#.

Explanation:

As a first step, if we just look at the equation we can see that the #x^2# term has a negative sign, so this parabola will be 'upside down', with the closed end at the top. The fact that there is a #3# in front of the #x^2# term means the parabola will be quite 'skinny'.

The #f(x)# intercept will occur when #x=0#, so:

#f(x)=-3(0^2)+0-5=-5#

The #x# intercept will occur when #f(x)=0# (similar to the #y# intercept).

#−3x^2+x−5=0#

I'd use the quadratic formula to solve this:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-1+-sqrt(1^2-4(-3)(-5)))/(2a)#

Help me out, colleagues: this yields imaginary roots. Not sure what the issue is. I'll work on the rest of the question but leave it with this query.

To find the vertex, we will use some calculus. The first derivative of the function, #f'(x)# describes the slope of the line:

#f'(x)=-6x+1#

This will be equal to zero at the vertex, because the slope goes to zero momentarily as it changes from positive to negative.

#-6x+1=0#

#-6x=-1#

#x=1/6#

To find the other coordinate of the vertex we substitute this into the expression for #f(x)#:

#f(x)=-3(1/6)^2+1/6-5=-4.75# (or #4 3/4# if you prefer).

Knowing the intercepts and the vertex, and that it is a narrow, upside down parabola, should be enough to draw a graph, but remember that you can also easily calculate a few additional points on the parabola. Substituting in points 1 unit either side of the #x# value of the vertex will yield information about the shape of the parabola, for example.