Question

Pls help me with these questions

Answer 1

Step-by-step explanation:

6 and 7 was already answered so I'll do 5 and 8.

5. The point the curve crosses the x axis is the x intercept, to find the x intercept of a rational function, we set y=0, and solve for x.

$0 = \frac{x - 4}{x}$

Set the numerator equal to 0.

$x - 4 = 0$

$x = 4$

So the x intercept is (4,0).

Next, to find gradient, we take the derivative of the function.

$\frac{x - 4}{x}$

We could use product rule, but for simplicity, serperate the function

$\frac{x}{x} - \frac{4}{x}$

$1 - \frac{4}{x}$

Next using exponents rules,

$1 - 4 {x}^{ - 1}$

Now we take the derivative,

Derivative of a constant is zero.

Derivative of a power function,

$n \times x {}^{n - 1}$

We move the exponent to the front, then we subtract the exponent by 1.

So, we get

$4 {x}^{ - 2}$

Now, we plug in. 4,

$4(4) {}^{ - 2}$

$4 \times \frac{1}{16} = \frac{1}{4}$

The slope or gradient at the x intercept is 1/4

8. The derivative of ax^2+bx, with respect to x is

$2ax + b$

When x=2, we have a gradient of 8.

$2a(2) + b = 8$

$4a + b = 8$

When x=-1, we have a gradient of -10.

$2a( - 1) + b = - 10$

$- 2a + b = - 10$

We have two system of equations,

$4a + b = 8$

$- 2a + b = - 10$

Let subtract the system to eliminate b.

$6a = 18$

$a = 3$

Plug 3 for a, back in to solve for b.

$4(3) + b = 8$

$12 + b = 8$

$b = - 4$

So a is 3

b is -4