Question

For the curve with equation y = "absmiddle" class="latex-formula"> + 4x + 3,find:
a)The gradient of the curve where it cuts the x-axis
b)the coordinates of the points on the curve where the tangent is:
i) parellel to the x-axis,
ii)parellel to the line 6x + 3y = 7.

Answer 1

Answer:

Below.

Step-by-step explanation:

a) First we find the derivative of the function, which gives us the gradient in terms of x:-

y = x^2 + 4x + 3

Gradient at x = dy/dx = 2x + 4.

When the curve cuts the x-axis, y = 0:

x^2 + 4x + 3 = 0

(x + 1)(x + 3) = 0

so x = -1 and -3  where the curve cuts the x axis.

The gradients at these point are therefore:

At x = -1 gradient = 2(-1) + 4 = 2

at x = -3 gradient = 2(-3) + 4 = -2.

b) (i) Where the tangent is parallel to x axis the gradient = 0

so 2x + 4 = 0 giving x = -2.

We now need to substitute x = -2 into the original function to find the y coordinate.

So the coordinates are (-2, (-2)^2 + 4(-2) +3))

= (-2, -1)

(ii) First find the slope of the line 6x + 3y =  7:

3y = -6x + 7

y = -2x + 7/3.

So the slope is -2.

Thus -2 = 2x + 4

2x = -6

x = -3

So the coordinates are (-3, (-3)^2 + 4(-3) + 3)

= (-3, 0).