A parabola has a vertex at #(5, 4)# and passes through #(6, 13/4)#. What are the x-intercepts?

1 Answer
Mar 2, 2017

Answer:

The x-intercepts are given by

#x = 5 +- (4sqrt(3))/3#

Explanation:

Start by finding the equation of the parabola. The vertex form of a parabola, with vertex #(p, q)#, is given by

#y = a(x- p)^2 + q#

We know an x-value, a y-value and the vertex. We can therefore set up an equation and solve for #a#.

#13/4 = a(6 - 5)^2 + 4#

#13/4 = a(1)^2 + 4#

#-3/4 = a#

The equation is therefore

#y = -3/4(x - 5)^2 + 4#

We can solve for the x-intercepts by taking the square root. Set #y# to #0#.

#0 = -3/4(x - 5)^2 + 4#

#-4 = -3/4(x - 5)^2#

#-4/(-3/4) = (x - 5)^2#

#16/3 = (x - 5)^2#

#+-4/sqrt(3) = x - 5#

#x = 5 +- 4/sqrt(3)#

#x = 5 +- (4sqrt(3))/3#

A graphical depiction of the parabola confirms our findings.

graph{y = -3/4(x - 5)^2 + 4 [-10, 10, -5, 5]}

Hopefully this helps!