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Tutorial on Quadratic Functions (1)

This is a tutorial on quadratic functions. The solutions and the explanations are detailed.

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:

  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)

  2. The graph of f passes through the point (2 , -8), it follows that
    f(2) = -8

  3. which leads to
    -8 = a(2 - 1)(2 + 2)

  4. expand the right side of the above equation and group like terms
    -8 = 4a

  5. Solve the above equation for a to obtain
    a = -2

  6. The equation of f is given by
    f(x) = -2(x - 1)(x + 2)

  7. 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:

  1. To find the points of intersection, you need to solve the system of equations
    y = x2 + x + 1

    y = mx

  2. Substitute mx for y in the first equation to obtain
    mx = x2 + x + 1

  3. Write the above quadratic equation in standard form.
    x2 + x(1 - m) + 1 = 0

  4. Find the discrimant D of the above equation.
    D = (1-m)2 - 4(1)(1)

    D = (1-m)2 - 4

  5. 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

  6. Solve the above inequality to obtain solution set for m in the intervals
    (-infinity , -1) U (3 , +infinity)

  7. 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

  8. Solve the above equation to obtain 2 solutions for m.
    m = -1

    m = 3

  9. 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

  10. 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.

    graphical solution to check


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.





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