Draw a diagram to show that there are two tangent lines to the parabola y=x^2 that pass through the point (0,-4). Find the coordinates of the points where these tangent lines intersect the parabola.

So far I have taken the derivative and got y'=2x..

Using point slope form I got y=2x-4....

I am not sure what to do next...

BUT, the given point is not on the curve, so what you did is bogus.

look at your sketch, the point (0,-4) is below the parabola.
Draw the two tangents, you will see that it touches the curve at two different points, one in the first quad, the other in the 2nd quad.
Let the point of contact be (a,b)

since (a,b) lies on the parabola, b = a^2
alsy s' = 2x = the slope at any point on the parabola
so at (a,b) the slope is 2a
but using the grade 9 way of finding slope
= (b+4)/(a-0) = (b+4)/a

so (b+4)/a = 2a
b+4 = 2a^2
b = 2a^2 - 4, but b = a^2
so 2a^2 - 4 = a^2
a^2 = 4
a = ± 2
then b = 4

so the two points of contact are (2,4) and (-2,4)

look at your sketch to see the validity of this answer
A further part of this question would be to find the equations of these two tangents.

To find the tangent lines to the parabola y = x^2 that pass through the point (0, -4), we can solve the system of equations formed by the equation of the parabola and the equation of the tangent lines.

Using the point-slope form, as you correctly found, the equation of the first tangent line is y = 2x - 4.

To find the coordinates of the points where this tangent line intersects the parabola, we can substitute y into the equation of the parabola:

x^2 = 2x - 4

Rearranging the equation:

x^2 - 2x + 4 = 0

To solve this quadratic equation, you can either complete the square or use the quadratic formula.

Completing the square:

x^2 - 2x + 1 + 4 = 1
(x - 1)^2 = -3

This equation has no real solutions, which means the first tangent line does not intersect the parabola.

Now, let's find the equation of the second tangent line:

Since both tangent lines pass through the point (0, -4), we can use the point-slope form again:

y + 4 = mx

To find the slope (m), we can differentiate the equation of the parabola:

y = x^2
y' = 2x

Substituting the x-coordinate of the point (0, -4) into the derivative:

-4 = 2(0)
-4 = 0

Since the slope is 0, the equation of the second tangent line is y = -4.

Substituting y into the equation of the parabola:

-4 = x^2

Solving for x:

x^2 = -4

This equation also has no real solutions, which means the second tangent line does not intersect the parabola either.

Therefore, there are no tangent lines to the parabola y = x^2 that pass through the point (0, -4).

To draw a diagram that shows two tangent lines to the parabola y = x^2 passing through the point (0, -4), we need to find the coordinates of the points where these tangent lines intersect the parabola.

You've correctly found the derivative of the parabola, which is y' = 2x. This represents the slope of a tangent line at any point on the parabola.

To find the tangent lines passing through the point (0, -4), we can use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

For the first tangent line, substituting the given point (0, -4) and the slope m = 2x, we have:
y - (-4) = 2(0 - x)
y + 4 = -2x
y = -2x - 4

For the second tangent line, substituting the same point (0, -4) and the negative slope m = -2x, we have:
y - (-4) = -2(0 - x)
y + 4 = 2x
y = 2x - 4

To find the coordinates of the points where these tangent lines intersect the parabola, we need to solve the system of equations formed by the parabola and each tangent line.

Substituting y = x^2 into the first tangent line equation, we have:
x^2 = -2x - 4
Rearranging, we get:
x^2 + 2x + 4 = 0

This is a quadratic equation, and we can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 2, and c = 4. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4*1*4)) / (2*1)
x = (-2 ± √(-12)) / 2
x = (-2 ± 2√3i) / 2
x = -1 ± √3i

Since the coefficients of the quadratic equation are real and the discriminant (√(-12)) is imaginary, the quadratic does not intersect the parabola at any real points. Hence, there are no points of intersection for the first tangent line.

Similarly, substituting y = x^2 into the second tangent line equation, we have:
x^2 = 2x - 4
Rearranging, we get:
x^2 - 2x + 4 = 0

Again, this is a quadratic equation, and we can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = -2, and c = 4. Plugging these values into the quadratic formula, we get:
x = (2 ± √((-2)^2 - 4*1*4)) / (2*1)
x = (2 ± √(4 - 16)) / 2
x = (2 ± √(-12)) / 2
x = 1 ± √3i

Again, since the coefficients of the quadratic equation are real and the discriminant (√(-12)) is imaginary, the quadratic does not intersect the parabola at any real points. Hence, there are no points of intersection for the second tangent line.

Therefore, there are no real points of intersection between the tangent lines and the parabola y = x^2.