Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18? I know x=-4 but I do not know that helps us. Please explain in steps!

The temperature of a point (x,y) in the plane is given by the expression x^2-y^2-4x+2y. What is the temperature of the coldest point on the plane? I know we can factor this to x(x-4)-y(y-2) but I am not sure what to do from there.

Thank you!

you know the minimum value is midway between the roots, which are -4 and 2p.

So, the vertex is at x = (-4+2p)/2 = p-2

So, now we want the y-coordinate of the vertex, which is

y = 2(p-2+4)(p-2-2p) = 2(p+2)(-p-2)
= -2(p+2)^2

If that is -18, then
-2(p+2)^2 = -18
p+2 = ±3, so
p = -1 or 5

You can plug those is and verify that the vertex is at y = -18

Note that

f(x,y) = (x^2-4x) + (y^2+2y)
= (x^2-4x+4) + (y^2+2y+1) - (4+1)
= (x-2)^2 + (y+1)^2 - 5

Clearly the minimum is achieved at (2,-1) where f(2,-1) = -5

For all other values of x and y, the first terms are positive.

To find all values of p such that 2(x+4)(x-2p) has a minimum value of -18, we can follow these steps:

Step 1: Expand the expression
2(x+4)(x-2p) = 2x^2 - 4px + 8x - 16p

Step 2: Simplify the expression
2x^2 - 4px + 8x - 16p = 2x^2 + (8 - 4p)x - 16p

Step 3: Identify the quadratic equation
We can see that this is a quadratic equation of the form ax^2 + bx + c. In this case, a = 2, b = 8 - 4p, and c = -16p.

Step 4: Find the vertex of the parabola
The vertex of a parabola can be found using the formula x = -b/(2a). By substituting the values, we get:
x = -(8 - 4p)/(2*2) = -(8 - 4p)/4 = (4p - 8)/4 = p - 2

Step 5: Substitute x = -4 into the equation
Since x = -4, we can substitute it into the quadratic equation:
2(-4)^2 + (8 - 4p)(-4) - 16p = -18

Step 6: Solve for p
16 + (-32 + 16p) - 16p = -18
16 - 32 + 16p - 16p = -18
-16 = -18

This equation is not true, which means there is no value of p that satisfies the condition. Therefore, there are no values of p that make the expression have a minimum value of -18.

Now, for the temperature of the coldest point on the plane:

Step 1: Given the expression x^2-y^2-4x+2y, we can rewrite it as x(x-4)-y(y-2).

Step 2: From the rewritten expression, we can observe that x is multiplied with x-4 and y is multiplied with y-2. This suggests that the coldest point would be obtained when x-4 = 0 and y-2 = 0.

Step 3: Solve for x and y
x-4 = 0 gives x = 4, and y-2 = 0 gives y = 2.

Step 4: Substitute the values of x and y into the original expression to determine the temperature at the coldest point:
Temperature = 4(4)-2(2) = 16-4 = 12.

Therefore, the temperature of the coldest point on the plane is 12.

To find all values of p such that 2(x+4)(x-2p) has a minimum value of -18, we can follow these steps:

Step 1: Expand the expression 2(x+4)(x-2p) to get 2x^2 - 4px + 8x - 16p.

Step 2: Simplify the expression to 2x^2 + (8 - 4p)x - 16p.

Step 3: We are looking for the minimum value of this quadratic expression. Since the coefficient of x^2 is positive, this quadratic opens upwards and will have a minimum value.

Step 4: The x-coordinate of the minimum point of a quadratic of the form ax^2 + bx + c is given by x = -b / (2a). In this case, a = 2, and b = 8 - 4p.

Step 5: Substitute the value of x = -4 into the equation x = -b / (2a). So, -4 = -(8 - 4p) / (2 * 2).

Step 6: Solve the equation for p. Multiply both sides by 4 to eliminate the fraction, which gives -16 = -(8 - 4p).

Step 7: Distribute the negative sign, resulting in -16 = -8 + 4p.

Step 8: Simplify the equation further to -8 = 4p.

Step 9: Divide both sides by 4 to solve for p. This gives p = -8 / 4, which simplifies to p = -2.

Hence, the value of p is -2.

For the temperature of a point (x, y), given by the expression x^2 - y^2 - 4x + 2y, we can follow these steps to find the coldest point on the plane:

Step 1: Rewrite the expression x^2 - y^2 - 4x + 2y as x(x - 4) - y(y - 2).

Step 2: Notice that this expression can be factored further. Rewrite it as (x - 4)(x - y)(y - 2).

Step 3: The temperature at any point (x, y) is given by this expression. To find the coldest point, we need to find the minimum temperature.

Step 4: Since the quadratic expression x(x - 4)(y - 2) can take on any real value, there is no minimum or maximum temperature in this case. Each point on the plane will have a different temperature.

Therefore, there is no coldest point on the plane.