Distance, Circles, And Quadratic Equations Worksheets Page 7
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9Appendix H: Distance, Circles, and Quadratic Equations
H7
From Figure H.9 we see that the parabola y = −x
+ 4x − 5 has no
2
Example 11
x-intercepts. This can also be seen algebraically by solving for the x-intercepts. Setting
y = 0 and solving the resulting equation
s
−x
+ 4x − 5 = 0
2
by the quadratic formula yields
√
−4 ±
16 − 20
Initial velocity
y =
= 2 ± i
−2
24.5 mls
0
Because the solutions are not real numbers, there are no x-intercepts.
Earth surface
A ball is thrown straight up from the surface of the Earth at time t = 0 s
Example 12
(a)
with an initial velocity of 24.5 m / s (Figure H.10a). If air resistance is ignored, it can be
s
shown that the distance s (in meters) of the ball above the ground after t seconds is given
by
s = 24.5t − 4.9t
40
2
(9)
30
(a) Graph s versus t, making the t-axis horizontal and the s-axis vertical.
20
(b) How high does the ball rise above the ground?
10
t
Equation (9) is of form (7) with a = −4.9, b = 24.5, and c = 0, so by (8)
Solution (a).
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5
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the t-coordinate of the vertex is
Time (s)
24.5
b
t = −
= −
= 2.5 s
(b)
2a
2(−4.9)
Figure H.10
and consequently the s-coordinate of the vertex is
s = 24.5(2.5) − 4.9(2.5)
= 30.625 m
2
y
The factored form of (9) is
s = 4.9t (5 − t)
so the graph has t-intercepts t = 0 and t = 5. From the vertex and the intercepts we obtain
Vertex
−b
/
(2a)
the graph shown in Figure H.10b.
Solution (b).
From the s-coordinate of the vertex we deduce that the ball rises 30.625 m
x
above the ground.
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THE GRAPH of x = ay
+ by + c
x = ay
+ by + c
2
a > 0
If x and y are interchanged in (7), the resulting equation,
x = ay
+ by + c
2
y
is called a quadratic equation in y. The graph of such an equation is a parabola with its line
of symmetry parallel to the x-axis and its vertex at the point with y-coordinate y = −b / (2a)
(Figure H.11). Some problems relating to such equations appear in the exercises.
Vertex
−b
/
(2a)
x
x = ay
+ by + c
2
a < 0
Figure H.11
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