What is the the vertex of #y = 2x^2-6x #?
3 Answers
The vertex is at
Explanation:
You could do this by the method of completing the square to find vertex form. But we can also factorise.
The vertex lies on the line of symmetry which is exactly half-way between the two
The
The midpoint is at
Now use the value of
The vertex is at
The vertex occurs at
Explanation:
We have:
# y=2x^2-6x #
which is a quadratic expression, with a positive coefficient if
Method 1:
We can complete the square:
# y=2{x^2-3x} #
# \ \ =2{(x-3/2)^2-(3/2)^2} #
# \ \ =2{(x-3/2)^2-9/4} #
# \ \ =2(x-3/2)^2-9/2 #
In this form we not that the first term
# 2(x-3/2)^2 = 0# when#x=3/2#
Wit this value of
# y = -9/2#
Method 2:
We can find the roots of the equation and use the fact that the vertex occurs at the midpoint the roots (by symmetry of quadratics)
For the roots, we have:
# 2x^2-6x = 0#
# :. 2x(x-3) = 0#
# :. x=0, x=3#
And so the midpoint (the
# x=(0+3)/2 = 3/2# , (as before).
And we find the
# y = 2(3/2)^2-6(3/2) #
# \ \ = 2 * 9/4 -6 * 3/2 #
# \ \ = 18/4-18/2 #
# \ \ = -18/4 #
# \ \ = -9/2 # , (as before)
We can verify these results graphically:
graph{y=2x^2-6x [-10, 10, -5, 5]}
vertex is at (1.5,-4.5)
Explanation:
So this is x intercept form we can easily find the x values when y is equal to zero.
We know that when we multiply if either product is zero the whole thing is zero.
So
and
So we know that x can be either 0 or 3 when y is zero.
We know that a parabola is symmetrical so half way between these points we will find the x value of the vertex.
So this is
So 1.5 is the x co-ordinate of the vertex so put in into the function to get the y co-ordinate
vertex is at (1.5,-4.5)