How to find the vertex of y = 2x^2 - 7? Pls teach me. thanks
do u mean the turning point?
You can either differentitate and set it to zero to find x
sub x back to the original equation to find y
or
complete the square
can i use x = -b/2a to find the vertex?
sounds good, can also
Originally posted by kittyslut:can i use x = -b/2a to find the vertex?
Quadratic Function
y = ax^2 + bx + c
Differentiate it
dy/dx = 2ax + b
For stationary point , 2ax + b = 0
x = - b / (2a)
If a > 0, substitute x = - b / (2a) into y to find the min y value
If a < 0, substitute x = - b / (2a) into y to find the max y value
This method of finding max y or min y value is not taught in "O" level maths textbooks.
But this is a very useful method that should be taught to "O" level maths students especially "O" level E.Maths students who do not learn differentiation ie to teach them that
For the quadratic function, y = ax^2 + bx + c
If a > 0, substitute x = - b / (2a) into y to find the min y value
If a < 0, substitute x = - b / (2a) into y to find the max y value
PS : This method is frequently used by China students.
Hi
one can easily derive the x value from the middle of the x values of the 2 roots as well :)
Originally posted by eagle:Hi
one can easily derive the x value from the middle of the x values of the 2 roots as well :)
Indeed, for drawing the turning point in a curve (especially in drawing a smooth curve question), many E.Maths teachers simply tell their students to find the middle value x of the 2 x-intercepts and then substitute this middle x value into the y to find the min y or max y.