Name: Non-linear Graphs
Uploaded: 2019-07-19
Duration: 4 min 41 s
Description: Here's an introduction to the graphs of four non-linear functions: quadratic, cubic, square root, and absolute value. Let us watch this video and understand the concept.

Question

Find the co-ordinates of the point at which the curve with equation $y = 2 x^{3} + 3 x^{2}$ has gradient $12$ .

Accepted Answer

We have to find the co-ordinates of the point at which the gradient of the curve with equation $y = 2 x^{3} + 3 x^{2}$ has gradient $18$ .

The derived function gives the gradient at the point where the $x$ co-ordinate is $x$ .

We know that,

If $y = a x^{n}$ , then $\frac{d y}{d x} = n \times a \times x^{n - 1}$

Now, $y = 2 x^{3} + 3 x^{2}$ and

$\frac{d y}{d x} = 3 \times 2 \times x^{3 - 1} + 2 \times 3 \times x^{2 - 1} = 6 x^{2} + 6 x$

Given that, gradient $\frac{d y}{d x} = 12$

$\Rightarrow 6 x^{2} + 6 x = 12$

$\Rightarrow x^{2} + x - 2 = 0$

$\Rightarrow x^{2} + 2 x - x - 2 = 0$

$\Rightarrow x (x + 2) - 1 (x + 2) = 0$

$\Rightarrow (x + 2) (x - 1) = 0$

$\Rightarrow x = - 2, x = 1$

Substituting $x = - 2$ , in the equation of the curve, we get,

$y = 2 {(- 2)}^{3} + 3 \times {(- 2)}^{2} = - 16 + 12 = - 4$

Substituting $x = 1$ , in the equation of the curve, we get,

$y = 2 {(1)}^{3} + 3 \times {(1)}^{2} = 2 + 3 = 5$

Hence, the co-ordinates of the point are $(- 2, - 4)$ and $(1, 5)$ .

Find the co-ordinates of the point at which the curve with equation $y = 2 x^{3} + 3 x^{2}$ has gradient $12$ .

Important Questions on Curved Graphs

Learn and Prepare for any exam you want !

Find the co-ordinates of the point at which the curve with equation y=2x3+3x2 has gradient 12.

Important Questions on Curved Graphs

Learn and Prepare for any exam you want !

Find the co-ordinates of the point at which the curve with equation $y = 2 x^{3} + 3 x^{2}$ has gradient $12$ .