1. ## function with derivative

Right I was given this problem to solve. Some of the questions I can do, others i am stuck with and need a maths guru to help.

So here goes, I am given on a grid the drawing of the graph y = xsquared - 2x -4

1 - I am asked to write down the coordinates of the minimum point of the curve - THIS I CAN DO, NO HELP NEEDED

2 - Use the graph to find estimates of the solutions to the equation xsquared - 2x - 4 = 0. Give your answer to 1d.p. THIS I CAN DO, NO HELP NEED

3 - Draw a suitable straight line on the grid to find estimates of the solutions of the equation xsquared - 3x - 6 = 0 THIS I CANNOT DO, AS IT IS A QUADRATIC AND THEREFORE WOULD PRODUCE A BELL SHAPE CURVE AND NOT A STRAIGHT LINE AS THEY SAY, NEED HELP PLEASE

4 - For y = xsquared - 2x - 4

4a) Find dy/dx CAN DO THIS, NO HELP NEEDED

4b) find the gradient of the curve at the point where x = 6 NEED HELP WITH THIS PLEASE

Nat

2. Originally Posted by Natasha1
Right I was given this problem to solve. Some of the questions I can do, others i am stuck with and need a maths guru to help.

So here goes, I am given on a grid the drawing of the graph y = xsquared - 2x -4

1 - I am asked to write down the coordinates of the minimum point of the curve - THIS I CAN DO, NO HELP NEEDED

2 - Use the graph to find estimates of the solutions to the equation xsquared - 2x - 4 = 0. Give your answer to 1d.p. THIS I CAN DO, NO HELP NEED

3 - Draw a suitable straight line on the grid to find estimates of the solutions of the equation xsquared - 3x - 6 = 0 THIS I CANNOT DO, AS IT IS A QUADRATIC AND THEREFORE WOULD PRODUCE A BELL SHAPE CURVE AND NOT A STRAIGHT LINE AS THEY SAY, NEED HELP PLEASE

4 - For y = xsquared - 2x - 4

4a) Find dy/dx CAN DO THIS, NO HELP NEEDED

4b) find the gradient of the curve at the point where x = 6 NEED HELP WITH THIS PLEASE

Nat
For (3)

$x^2-3x-6=\left(x^2-2x-4\right)-(x+2)$

This is zero where $x^2-2x-4=x+2$

Hence draw the line $y=x+2$ and find where it intersects $x^2-2x-4$

3. Not sure I fully understand this, how about 4b?

4. Originally Posted by Natasha1
Not sure I fully understand this, how about 4b?
$y=x+2$ is a line.

If you subtract the linear equation from the quadratic, you get the new quadratic.
Subtracting 2 things that are equal give zero, and they are equal at the points of intersection.

(4)

$\displaystyle\frac{dy}{dx}$ is the slope of the tangent to the curve (the gradient of the curve) over the domain of x.

So you first calculate the derivative (instantaneous gradient of the curve) which is $\displaystyle\frac{dy}{dx}$ and evaluate it for x=6
(the derivative will be in terms of x).

5. Ok so dy/dx = 2x - 2

but how do you evaluate it for x = 6, I have never done this. Just need clear explanation to hopefully understand.

6. Originally Posted by Natasha1
Ok so dy/dx = 2x - 2

but how do you evaluate it for x = 6, I have never done this. Just need clear explanation to hopefully understand.
Ok,

you have evaluated the derivative.
This is a formula for the gradient of the curve, which varies with x
(imagine the tangent to the curve changing slope as you move from one value of x to another along the x axis).

If you now place any real value of x into the derivative (formula for calculating the gradient),
you will evaluate the curve gradient at that value of x.

So, if you substitute x with 6 in 2x-2, to get 2(6)-2, that will be the curve gradient when x is 6.

7. So the gradient of the curve is 10

8. Originally Posted by Natasha1
So the gradient of the curve is 10
Yes, when x is 6 only.

9. Thanks for clearly explaining

10. You're welcome Natasha,

here is a graph of the situation

11. Here is a graph for part (3)