Quadratic functions/cubic functions
Quadratic Equations and Inequalities introduces students to the graphs of quadratics, teaches them how to find the vertex, intercepts, discriminant, domain and range and interpret the graph in relation to these qualities. It as well teaches the students the focus and directrix of parabolas, how to use a sign number line to obtain the shape of a graph
Quadratic functions: y = f(x) = ax2 + bx + c (a ≠ 0)
where a, b, c are real numbers.
The Graph of a Quadratic Function
Shape: parabola, symmetric
When a > 0
When a < 0
* Vertex:
(-b/2a, f(-b/2a))
f(-b/2a) is the optimum value (either maximum or minimum) of the function.
* Zeros of quadratic functions = Solutions of the quadratic equation ax2 + bx + c = 0
x = (-b ± (√ b2 - 4ac)/ 2a
* y-intercept means the value of y when x = 0
y = a(0)2 + b(0) + c = c
Examplef(x) = x2 + 2x - 3
Find
(a) The vertex
(b) Is the vertex a maximum or minimum point?
(c) x value for the optimal value
(d) The optimal value
(e) Zeros
(f) y-intercept
(g) Graph
Example 2
If 100 feet of fence is used to enclose a rectangular yard, determine the length and width of the rectangle that give maximum area.
Shifting Graphs
Compare vertices.
Average Rate of change
The rate of change of a quadratic function is not constant.
Clue: Make use of the average rate of change between two points.
The average rate of change of f(x) between two points x = a and x = b (a < b) is
Average rate of change
= Slope of the line between (a, f(a)) and (b, f(b))
General Polynomial Inequalities
Solve x5 + 3x4 – 23x3 – 51x2 + 94x + 120 > 0.
First, I factor to obtain the zeroes:
x5 + 3x4 – 23x3 – 51x2 + 94x + 120
= (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0
...therefore x = –5, –3, –1, 2, and 4 are the zeroes of this polynomial.
To solve by the Test-Point Method, you would select a sample point in each interval, the intervals would be (negative infinity, –5), (–5, –3), (–3, –1), (–1, 2), (2, 4), and (4, positive infinity). As is evident, if your polynomial or rational function possesses a lot of factors, the Test-Point Method can become quite time-consuming.
To solve by the Factor Method, you would solve each factor for its positivity: x + 5 > 0 for x > –5; x + 3 > 0 for x > –3; x + 1 > 0 for x > –1; x – 2 > 0 for x > 2; and x – 4 > 0 for x > 4.
Rational Inequalities
Solve x/(x – 3) < 2.
To start with, you must remember that you can't start solving until you have the inequality in "= 0" format.
x /x - 3 ≤ 2
After this, you need to convert to a common denominator:
(x /x - 3) - (2(x - 3) / x - 3) ≤ 0
The two factors are –x + 6 and x – 3. Observe that x cannot equal 3, or else you would be dividing by zero, which is not permitted. The first factor, –x + 6, equals zero when x = 6. The other factor, x – 3, equals zero when x = 3. Now, x cannot in reality be equal to 3, therefore, this endpoint will not be added in any solution interval (even though this is an "or equal to" inequality), but you require the value in order to find out what your intervals are. In this situation, my intervals are (negative infinity, 3), (3, 6], and [6, positive infinity). Observe the use of brackets to show that 6 can be included in the solution, but that 3 cannot.
With the use of the Test-Point Method, you would pick a point in each interval and test for the sign on the result. You could use, for example, x = 0, x = 4, and x = 7.
Making use of the Factor Method, you solve for each factor: –x + 6 > 0 for –x > –6, or x < 6; x – 3 > 0 for x > 3.
Therefore, the solution is all x's in the intervals (negative infinity, 3) and [6, positive infinity).
There is still another way to solve inequalities. You would still need to find the zeroes (x-intercepts) first, but then you graph the function, and just look: wherever the graph is above the x-axis, the function is positive; wherever it is below the axis, the function is negative. For instance, for the first quadratic exercise, y = x2 – 3x + 2 > 0, we found the zeroes at x = 1 and x = 2.
You would choose the two intervals (but not the interval at endpoints) where the line is above the x-axis.
Or that gigantic polynomial we did earlier: x5 + 3x4 – 23x3 – 51x2 + 94x + 120 > 0. We found the zeroes of the polynomial, being x = –5, x = –3, x = –1, x = 2, and x = 4. See the graph below:
A graphing calculator can save you a lot of time on these inequalities. You ought to still show your work and reasoning, but ensure you make use of pictures to confirm the algebra.
Solving Quadratic Inequalities: Examples
Solve 2x2 + 4x > x2 – x – 6.
The two connected two-variable equations in this case are y = 2x2 + 4x and y = x2 – x – 6.
This inequality is asking when the parabola for y = 2x2 + 4x (in green) is higher than the parabola for y = x2 – x – 6 (in blue):
You would observe that it is difficult to tell where the green line (y = 2x2 + 4x) is above the blue line
(y = x2 – x – 6). Therefore, instead of trying to solve this inequality, it is better to work with the related inequality below:
2x2 + 4x > x2 – x – 6
2x2 + 4x – x2 + x + 6 > 0
x2 + 5x + 6 > 0
This last inequality is simpler to because, all you need to do is find the zeroes of y = x2 + 5x + 6 (which is easy) and then choose the correct intervals based on just the one parabola (which is as well is not difficult). That is, it is easier to compare one parabola with the x-axis than to compare two parabolas with each other. But because the one parabola (y = x2 + 5x + 6) came from the linkage or amalgamation of the two original parabolas ("paraboli"?), the solution to the simpler one-parabola inequality will be similar to the solution of the original two-parabola inequality. Due to the fact that the solutions will be the same, you'll choose to work with the simpler case.
The equation: "2x2 + 4x > x2 – x – 6" was simplified to obtain "x2 + 5x + 6 > 0". The related two-variable equation is y = x2 + 5x + 6. To start with, you will find the zeroes (that is, the x-intercepts):
x2 + 5x + 6 = 0
(x + 2)(x + 3) = 0
x = –2 or x = –3
These two intercepts split the number-line into three intervals, namely x < –3, –3 < x < –2, and x > –2. On which of these three intervals is y = x2 + 5x + 6 above the x-axis? Since y = x2 + 5x + 6 graphs as a right-side-up parabola, the quadratic is above the axis on the ends:
Then the solution is:x < –3 or x > –2
Why was this solution "or equal to", instead of merely "greater than" or "less than"? This is so because the intial inequality was "or equal to", so the boundary points, being the zeroes or x-intercepts, are included in the solution. Copts Reserved
Solve x2 + x + 1 > 0.
For this particular inequality, you would calculate the zeroes:
Since there is a negative inside the square root, there ought not to be any x-intercepts. That is, this quadratic must be either always above the x-axis or else constantly below, because it can never cross or touch the axis.
Since y = x2 + x + 1 is a "positive" quadratic, the parabola is right-side-up, therefore, you known that it would go up indefinitely. For the parabola not to cross the x-axis, it must be that the parabola is always above the axis, as you could observe from this graph:
Therefore, when is y = x2 + x + 1 greater than zero (above the axis)? Constantly. Therefore the answer to the question is:
The above solution could as well be stated as "all real numbers" or written as the interval "from negative infinity to positive infinity".
Solve x2 + x + 1 < 0.
This looks similar to the just finished question, apart from the fact that in this present case what you would find is where the parabola is below the axis. You know already that there are no x-intercepts. Again, due to this, it is a right-side-up parabola. The graph is constantly above the axis. Therefore, where is y = x2 + x + 1 less than zero? Nowhere! Thus the answer is:
View more on psalmfresh.blogspot.com for more educating and tricks article Thank you
No comments
Post a Comment