Revision notes on Quadratic Graphs for Edexcel AS Maths.

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• Created by: Anonymous
• Created on: 15-10-07 11:01

First 228 words of the document:

y = x 2x15
2
The vertex The turning point
This can be the maximum or minimum point of the graph. It can be found either by
completing the square:
y = (x1)216
vertex = (1.16)
or by differentiating f(x), because the gradient at the vertex is 0 and then substituting the
value of x into the equation:
f'(x) = 2x2
2x2 = 0
x=1
y = 122x115 = 16
vertex = (1,16)
The tangent A line that touches the curve
A tangent has the same gradient as the curve at the point where it touches. The equation for
a tangent is found by differentiating f(x) to find the gradient of the curve, then substituting in
the xcoordinate of the point where the tangent touches the curve. This gives the gradient of
the tangent. Ex: (3,12)
f'(x) = 2x2
f' (3) = 4
y+12 = 4(x3)
The xintercepts The points where the curve crosses the xaxis
The xintercepts can be found by substituting y=0 into the equation and then solving the
equation to find the values of x:
x22x15 = 0
(x+3)(x5) = 0
x= 3 or 5
The yintercepts The point where the curve crosses the yaxis
This can be found by substituting x=0 into the equation to find the value of y