graph the quadratic function y=3x-x^2
Answers
I wanted to give a quick derivation of why the vertex point is at
x
=
−
b
2
a
for anyone who understands differential calculus:
The vertext point is also know as the stationary point, where the functions gradient is
0
Let
y
=
a
x
2
+
b
x
+
c
Using power rule:
d
y
d
x
=
2
a
x
+
b
When gradient is 0:
2
a
x
+
b
=
0
⇒
2
a
x
=
−
b
⇒
x
=
−
b
2
a
The quadratic equation x² – 3x – 10 = 0 can be solved by any one of the following four methods:
(1.) Factoring
The quadratic expression in x can be easily factored as follows:
x² – 3x – 10 = (x + 2)(x – 5) = 0
Now, by the Zero Product Property, i.e., if “a” and “b” are real numbers, then ab = 0 if and only if a = 0 or b = 0; therefore,
x + 2 = 0 or x – 5 = 0
x + 2 – 2 = 0 – 2 or x – 5 + 5 = 0 + 5
x = –2 or x = 5
Therefore, the solution set of our given quadratic equation is {–2, 5}.
(2.) The Quadratic Formula
The Quadratic Formula which is used to solve any quadratic equations is derived from the general quadratic equation, i.e., ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0, and is as follows:
x = [–b ± √(b² – 4ac)]/2a
Now, substituting into this formula for the given quadratic equation, x² – 3x – 10 = 0, where a = 1, b = –3, and c = –10, we have:
x = {–(–3) ± √[(–3)² – 4(1)(–10)]}/2(1)
= {–(–3) ± √[9 – (–40)]}/2
= {–(–3) ± √[9 + 40]}/2
= {3 ± √[49]}/2
= {3 ± 7}/2
x = {3 + 7}/2 or x = {3 – 7}/2
x = 10/2 = 5 or x = –4/2 = –2
Therefore, the solution set is {–2, 5}
(3.) Competing the Square
x² – 3x–10 = 0 (the coefficient on the x²-term must be 1; otherwise, an adjustment needs to be made to make it equal 1)
x² – 3x – 10 + 10 = 0 + 10 (We need a perfect square expression on
the left side, but –10 does not give us
a perfect square expression on the left,
so add 10 to both sides)
x² – 3x + (3/2)² = 10 + (3/2)² (Take one-half the coefficient of
the x-term and square the
result to get the required constant
term on the left. Also, add this term
on the right)
x² – 3x + (9/4) = 10 + (9/4)
x² – 3x + (9/4) = 49/4
Now, we have a perfect square trinomial on the left, and we can express it as follows:
[x – (3/2)]² = 49/4
Now, take the square root of both sides, we get:
x – (3/2) = ±√(49/4)
x – (3/2) = ±7/2
Now, solving the two equations for x, we get:
x – (3/2) = 7/2 or x – (3/2) = –7/2
x – (3/2) + (3/2) = (7/2) + (3/2) or x – (3/2) + (3/2) = –7/2 + (3/2)
x = 10/2 or x = –4/2
x = 5 or x = –2
Therefore, the solution set is {–2, 5}
(4.) Graphing
Using a graphing calculator or a sheet of graph paper on which you can construct a Cartesian or rectangular (xy) coordinate system, plot the given quadratic expression x² – 3x – 10 as a function of x, i.e., y = f(x) = x² – 3x – 10. To do this plot, plug in a convenient range of x-values, e.g., x = –3 to x = 6 in 0.5 increments, into x² – 3x – 10 to get the corresponding y-values and then plot the resultant points (x,y) to obtain a good picture of the resultant parabolic graph. You will find that the graph (parabola) crosses the x-axis at the points (–2, 0) and (5, 0) which specify the x-values, –2 and 5, which, in turn, satisfy the equation y = f(x) = x² – 3x–10 = 0.
Checking the solutions, x = –2 and x = 5, for x² – 3x – 10 = 0:
x = –2 x = 5
x² – 3x – 10 = 0 x² – 3x – 10 = 0
(–2)² – 3(–2) – 10 = 0 (5)² – 3(5) – 10 = 0
4 – (–6) – 10 = 0 25 – 15 – 10 = 0
4 + 6 – 10 = 0 25 – 15 – 10 = 0
10 – 10 = 0 10 – 10 = 0
0 = 0 0 = 0