How do i solve the pair of simultaneous equations
y=x^2-3x-12 and y=2x-16
and epxress y=x^2-3x-12 in the form y=(x-h^2)+k, where h and k are constants
Hence determine the minimum value of the function y=x^2-3x-12
I have a few of these to do, so if someone can just show me how to do this one i can do the rest.
in the first:
notice that one equation is quadratic, the other is linear.
sub the linear into the quadratic
x^2 - 3x - 12 = y = 2x - 16
x^2 - 3x - 12 = 2x - 16
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or x = 4
sub into the easier of the two, namely into y = 2x - 16
if x = 1 then y = 2(1) - 16 = -14
if x = 4 then y = 2(4) - 16 = -8
so two points of intersection, (1,-14) and (4,-8)
We were lucky that the quadratic factored, if not you will have to use the quadratic equation formula.
for the second, complete the square
y=x^2-3x-12
y = x^2 - 3x + 9/4 - 9/4 - 12
= (x - 3/2)^2 - 57/4
so the min value of the function is -57/4 and it occurs when x = 3/2
or
the vertex is (3/2,-57/4)
Here a link to the graph of this problem:
http://i263.photobucket.com/albums/ii157/mathmate/raquel.png
To solve the pair of simultaneous equations y = x² - 3x - 12 and y = 2x - 16, we can equate both equations and solve for x.
By setting x² - 3x - 12 equal to 2x - 16:
x² - 3x - 12 = 2x - 16
Rearranging the equation:
x² - 3x - 2x - 12 + 16 = 0
x² - 5x + 4 = 0
Now we can solve this quadratic equation by factoring or by using the quadratic formula.
Factoring:
(x - 4)(x - 1) = 0
Setting each factor equal to zero:
x - 4 = 0 or x - 1 = 0
Solving for x:
x = 4 or x = 1
Now that we have found the values of x, we can substitute them back into one of the original equations to find the corresponding values of y.
Using the equation y = x² - 3x - 12:
For x = 4: y = (4)² - 3(4) - 12 = 16 - 12 - 12 = -8
For x = 1: y = (1)² - 3(1) - 12 = 1 - 3 - 12 = -14
Therefore, the solutions to the simultaneous equations are (4, -8) and (1, -14).
To express y = x² - 3x - 12 in the form y = (x - h)² + k, we need to complete the square.
y = x² - 3x - 12
= (x² - 3x + 9/4 - 9/4) - 12 [Adding and subtracting (9/4)]
= (x² - 3x + 9/4) - 9/4 - 12
= (x - 3/2)² - 9/4 - 48/4
= (x - 3/2)² - 57/4
Therefore, y = (x - 3/2)² - 57/4 is the expression of y in the given form, where h = 3/2 and k = -57/4.
Finally, to determine the minimum value of the function y = x² - 3x - 12, we look at the coefficient of the squared term in the expression: (x - 3/2)².
Since the coefficient is positive, the parabola opens upwards, and hence, the minimum value occurs at the vertex.
The x-coordinate of the vertex can be found using the formula: x = -b/2a, where a, b, and c are the coefficients of x², x, and the constant term in the quadratic equation, respectively.
For the given equation: y = x² - 3x - 12, a = 1 and b = -3.
x = -(-3)/2(1) = 3/2 = 1.5
Substituting the value of x back into the equation:
y = (1.5 - 3/2)² - 57/4
= (-0.5)² - 57/4
= 0.25 - 57/4
= 0.25 - 14.25
= -14
Therefore, the minimum value of the function y = x² - 3x - 12 is -14.
To solve the pair of simultaneous equations, we can start by setting the two expressions for y equal to each other:
x^2 - 3x - 12 = 2x - 16
Now we have a quadratic equation. We can rearrange it to be in standard form:
x^2 - 3x - 12 - 2x + 16 = 0
x^2 - 5x + 4 = 0
Now we can factorize the quadratic equation:
(x - 4)(x - 1) = 0
This gives us two possible solutions for x:
x - 4 = 0 --> x = 4
x - 1 = 0 --> x = 1
Now that we have the values of x, we can substitute them back into one of the original equations to find the corresponding values of y. Let's use the second equation:
y = 2x - 16
For x = 4:
y = 2(4) - 16
y = 8 - 16
y = -8
For x = 1:
y = 2(1) - 16
y = 2 - 16
y = -14
So the solutions to the pair of simultaneous equations are (4, -8) and (1, -14).
To express y = x^2 - 3x - 12 in the form y = (x - h)^2 + k, we need to complete the square. Here's how you can do it:
1. Start with the expression y = x^2 - 3x - 12.
2. Group the terms involving x together: y = (x^2 - 3x) - 12.
3. To complete the square, take half of the coefficient of x (-3) and square it: (-3/2)^2 = 9/4.
4. Add and subtract this value inside the parentheses: y = (x^2 - 3x + 9/4 - 9/4) - 12.
5. Rearrange the terms: y = (x^2 - 3x + 9/4) - 9/4 - 48/4.
6. Simplify: y = (x - 3/2)^2 - 57/4.
Now, we have expressed y = x^2 - 3x - 12 in the form y = (x - h)^2 + k, where h = 3/2 and k = -57/4.
To determine the minimum value of the function y = x^2 - 3x - 12, we can observe that the minimum value occurs when x is equal to the x-coordinate of the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c.
In our case, the quadratic equation is y = x^2 - 3x - 12, so the coefficients are a = 1, b = -3, and c = -12.
Using the formula x = -b/2a, we can find the x-coordinate of the vertex:
x = -(-3)/(2*1) = 3/2
Now, substitute this value of x back into the equation to find the corresponding y-coordinate:
y = (3/2 - 3/2)^2 - 57/4
y = 0 - 57/4
y = -57/4
So, the minimum value of the function y = x^2 - 3x - 12 is -57/4.