Question

Pls help me with these questions

Answer 1

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 = (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.

`(x - 4)/(x)`

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

`(x)/(x) - (4)/(x)`

`1 - (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 * 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 * (1)/(16) = (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