How do you find the vertex and intercepts for #y = 3(x - 2)^2+1#?

1 Answer
iceman
Mar 17, 2016

The vertex is at:#(2, 1)#
There are no x-intercepts
The y-intercepts is at:#(0, 13)#

Explanation:

#y=a(x - h)^2+k# => Equation of the parabola in vertex form where the vertex is at:#(h, k)#, so in this case:
#y=3(x-2)^2+1# , the vertex is at:#(2, 1)#

To find the x-intercepts set y to zero and solve for x:
#3(x-2)^2+1=0#
#3(x-2)^2=-1#
#(x-2)^2=-1/3# => Since the square of a number can not be negative the roots of this quadratic equation are complex i.e: no real roots, which means the parabola does not cross the x-axis hence there are no x-intercepts.

To find the y-intercept set x to zero and solve for y:
#y=3(0-2)^2+1#
#y=3(-2)^2+1#
#y=3(4)+1#
#y=12+1#
#y=13#
Hence the y-intercepts is at:#(0, 13)#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1977 views around the world
You can reuse this answer
Creative Commons License