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This is a tutorial on quadratic functions. The solutions and the explanations are detailled.
Example 1 : Find the equation of the quadratic function f whose graph passes through the point (2 , -8) and has x intercepts at (1 , 0) and (-2 , 0).
Solution to Example 1:
- Because the graph has x intercepts at (1 , 0) and (-2 , 0), the equation of the function may be written as follows.
f(x) = a(x - 1)(x + 2)
- The graph of f passes through the point (2 , -8), it follows that
f(2) = -8
- which leads to
-8 = a(2 - 1)(2 + 2)
- expand the right side of the above equation and group like terms
-8 = 4a
- Solve the above equation for a to obtain
a = -2
- The equation of f is given by
f(x) = -2(x - 1)(x + 2)
- Check answer
f(1) = 0
f(-2) = 0
f(2) = -2(2 - 1)(2 + 2) = -8
Matched Exercise: Find the equation of the quadratic function f whose graph has x intercepts at (-1 , 0) and (3 , 0) and a y intercept at (0 , -4).
Answers to above exercise.
Example 2 : Find values of the parameter m so that the graph of the quadratic function f given by
f(x) = x2 + x + 1
and the graph of the line whose equation is given by
y = mx
have:
a) 2 points of intersection,
b) 1 point of intersection,
c) no points of intersection.
Solution to Example 2:
- To find the points of intersection, you need to solve the system of equations
y = x2 + x + 1
y = mx
- Substitute mx for y in the first equation to obtain
mx = x2 + x + 1
- Write the above quadratic equation in standard form.
x2 + x(1 - m) + 1 = 0
- Find the discrimant D of the above equation.
D = (1-m)2 - 4(1)(1)
D = (1-m)2 - 4
- For the graph of f and that of the line to have 2 points of intersection, D must be positive, which leads to
(1-m)2 - 4 > 0
- Solve the above inequality to obtain solution set for m in the intervals
(-infinity , -1) U (3 , +infinity)
- For the graph of f and that of the line to have 1 point of intersection, D must be zero, which leads to
(1-m)2 - 4 = 0
- Solve the above equation to obtain 2 solutions for m.
m = -1
m = 3
- For the graph of f and that of the line to have no points of intersection, D must be negative, which leads to
(1-m)2 - 4 < 0
- Solve the above inequality to obtain solution set for m in the interval
(-1 , 3)
The graphs of y = 3x, y = -x and that of the quadratic function are shown in the figure below.
Matched Exercise: Find values of the parameter c so that the graph of the quadratic function f given by
f(x) = x2 + x + c
and the graph of the line whose equation is given by
y = 2x
have:
a) 2 points of intersection,
b) 1 point of intersection,
c) no points of intersection.
Answers to above exercise.
More references and links on the quadratic functions in this website.
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