A curve is given by the equation: y = x^2 – 2x – 3?

Question

a)Evaluate y at x = -2, -1, 0 , 1, 2, 3, 4

b) according to that curve`s equation. What are the solutions to the following equations:

X^2 – 2x – 3 = 0
X^2 – 2x – 3 = x
X^3 – 2x – 3 = -2



c) what is the minimum value of x^2 – 2x – 3 ?

Answer 1

a) y = 5, 0, -3, -4, -3, 0, 5

b) -1 and 3
4.15 and -1.15
2.41 and -0.41

c) -4

Answer 2

As has been stated, part a is simple. In each case, replace x with the value and perform the arithmetic. The major mistake here is in squaring negative numbers - the result is positive, but people using calculators often forget parentheses. In general, be careful when minus signs are involved.

In part b, at least one solver used the quadratic formula. Here it is important to state what kind of problem this is.

The first equation can be solved simply by looking at the y-values in part a. When you substitute -1 and 3 for x, you get 0 back.
The second equation has irrational solutions, and may be solved using the quadratic formula or by completing the square:
x^2 - 2x - 3 = x
x^2 - 3x = 3
x^2 - 3x + (3/2)^2 = 3 + (3/2)^2
(x - 3/2)^2 = 3 + 9/4 = 21/4
x - 3/2 = +/- sqrt(21)/2
x = 3/2 +/ sqrt(21)/2
Quadratic formula: x = (-b +/- sqrt(b^2 - 4ac))/2a, with a = 1, b = -3, and c = -3, gives
x = (- - 3 +/- sqrt((-3)^2 - 4*1*(-3)))/(2*1) = (3 +/- sqrt(9 + 12))/2 = (3 +/- sqrt(21))/2.

The third is a cubic equation, unlike the first two, which are quadratic. So, it is possible that this is a misprint, since no other part of the problem refers to an expression with a cube. Generally cubic equations are not so easy to solve. The only possible rational solutions are +/-1, but x = -1 works. So, x^3 - 2x - 3 = -2 implies x^3 -2x -1 = 0, so
(x + 1)(x^2 - x - 1) = 0. Now, x^2 - x = 1, x^2 - x + 1/4 = 5/4, so (x - 1/2 = sqrt(5)/2, and x = (1 +/- sqrt(5))/2 are the other two solutions.
However, it is more likely that the actual third equation is x^2 - 2x - 3 = -2. In this case, x^2 - 2x = 1, so x^2 - 2x + 1 = 2, and (x - 1)^2 = sqrt(2), so x - 1 = +/-sqrt(2), and x = 1 +/- sqrt(2). Also, the quadratic formula can be used.

Each of these can be solved using the TI-83, and related calculators: To solve f(x) = g(x): Go to "Y=" and enter f(x) in Y1, g(x) in Y2. (If the right hand side is 0, enter 0 in Y2). Then graph. If you can see the intersection point(s) on the screen, then go to 2nd "trace", 5, enter, enter, then move the cursor close to one of the intersection points, and hit enter again. (Notice that of g(x) = 0, then the second curve is the x-axis.)
Using this approach on the cubic equation gives x = 1.618034, -0.618034, and -1. The first two values are rounded.

In c, the graph of a quadratic polynomial is a parabola. If the squared tern is poitive, the parabola opens upward and has a minimum at the vertex. If it is negative, the parabola opens downward and has a maximum at its vertex. The vertex can be found in several ways. (1) complete the square. x^2 - 2x - 3 = x^2 - 2x + 1 - 1 - 3 = (x - 1)^2 - 4. Thus, the minimum is
-4 when x = 1. (2) If the parabola intersects the x-axis in one point, that is the vertx, if it intersects in two points, their average is the x-ccodinate of the vertex. So, (-1 + 3)/2 = 2/2 = 1. The minimum value is y = (1)^2 - 2(1) - 3 =
-4. This fails if the parabola misses the x-axis. (3) using the calculator, go to "y=", x^2 - 2x - 3, graph, go to 2nd trace, and choose 3 You will need to go to the left of the min and hit enter, then to the right of the min and hit enter, and finally hit enter again. The y-coordinate is the minimum value. In this particular case, you can probably just read the minimum value right off graph. Also, way back in part a, it is clear that the y-values are the same on either side of x = 1, so given the simplicity of the curve, the bottom value should be -4.

sorry to be so long-winded. I don't know if this was supposed to be a calculor problem or what.

Answer 3

a) is too easy
y = x^2 – 2x – 3
x=-2 y= (-2)^2 -2*(-1) - 3 = 5
x=-1 y=0
x=0 y=-3
x=1 y=-4
x=2 y= -3
x=3 y= 0
x=4 y= 5


b:)
if ax^2 + bx+c=0 den

x= {-b+sqrt(b^2-4ac)}/2a and

x={-b-sqrt(b^2-4ac)}/2a

and now
a=1 b=-2 c=-3
a=1 b=-3 c=-3
i do not know how to solve X^3 – 2x – 3 = -2


c)

y= x^2 – 2x – 3 so y'= 2x-2=0 x=1
y= 1^2- 2*1 -3 =-4

min y= -4

Answer 4

y=x^2-2x-3=(x-1)^2-4
to find the roots, put y=0;
x-1=+/-2;
x=3,-1.
the minimum value of y=-4, when x-1=0

Answer 5

a) 5, 0, -3, -4, -3. -0, 5

b) x = -1 and x =3 [that's where y = 0]
x = [3+/- sqrt(21)]/2
x = -1 , x= [-1 +/- iSqrt(3)]/2

c) x = 0

Answer 6

It seems such as you like greater suggestion on the 2nd equation. is it tangent to Y or? as quickly as you have those you may draw them and use the integration formulation. in case you have a calculus e book by swokowski it rather is definition 6.a million. desire it facilitates!

Answer 7

im wondering if you couldnt do this on your own, this is no "do my homework" board or anything like this, this is like basics, if you cant do this on your own, then i wish you good luck in the future.