If you have covered differentiation (calculus), equate f'(x) = 0 to find the value of x through which passes the axis of symmetry. Substitute into f(x) to get the value of y. The solutions of f(x)=0 can be found by either factoring or the quadratic formula.
Without using calculus, a quadratic can be transformed into its canonical form:
f(x) = a(x-h)² + k
by completing the square.
x=h is the axis of symmetry which passes through the vertex.
The vertex is the point (h, f(h))
And the roots are
h ± sqrt(-k/a)
For the case in point, assuming the correct expression is 2x²-7x+5,
2x²-7x+5
=2(x² -(7/2)x) + 5
=2(x-7/4)² + 5 - 2(7/4)²
=2(x-7/4)² - 9/8
Thus a=2, h=-7/4, k=-9/8
The axis of symmetry is x=7/4
The vertex is (7/4, f(7/4)=-9/8)
The roots are
7/4 + sqrt(-(-9/8)/2) = 5/2
7/4 - sqrt(-(-9/8)/2) = 1
Try to reproduce the answer and thus provide a check of the calculations.
on a graph of 2x^2 -7+5 use equations to find the vertex point ,roots and asis of symmetry
2 answers
A sketch of the function is shown at the following link:
http://i263.photobucket.com/albums/ii157/mathmate/anonymous.png
http://i263.photobucket.com/albums/ii157/mathmate/anonymous.png
2 answers