Question

Find the co-ordinates of the point on graph y = 5x^2 − 3x + 1 where the gradient is 2

Answer 1

Answer:

(1/ 2, 3/ 4)

Step-by-step explanation:

• Gradient of a curve is the slope of the tangent tonthe curve at a point.

• To find gradient of a curve differentiate it's equation with respect to x.

Equation of the curve

y = 5x² - 3x + 1

differentiating with respect to x

dy/ dx = 2. 5x - 3

dy/ dx = 10x - 3

This derivative is the gradient and gradient has to be equal to 2.

10x - 3 = 2

10x = 5

x = 1/ 2

placing this value of x in the equation to get the y coordinate

y = 5(1/ 2)² - 3(1/ 2) + 1

y = 5/ 4 - 3/ 2 + 1

y = 3/ 4

Therefore, the point at which the gradient of the curve is 2 is

(1/ 2, 3/ 4)

Derivatives used :-

• d/ dx (x ^ n) = nx ^ (n -1)

• d/ dx (ax) = a

• d/ dx ( constant) = 0

________________

Hope this helps!

Answer 2

Given:

Equation of the curve, y = 5x² - 3x + 1

Gradient = 2

To find :

The co-ordinates of the point on graph y = 5x² - 3x + 1 where the gradient is 2.

Solution:

Step 1 of 2:

To find the coordinates of the point, differentiate the given equation with respect to x,

y = 5x² - 3x + 1

$\frac{dy}{dx}$ = 2 × (5)x = 3

Where,

$\frac{d}{dx} (x^{n} ) = nx^{n-1} \\\frac{d}{dx} (ax) = a \\\frac{d}{dx} (constant) = 0$

$\frac{dy}{dx}$ = 2 (5)x - 3 + 0

$\frac{dy}{dx}$ = 10x - 3

Step 2 of 2:

$\frac{dy}{dx}$ = 10x - 3 is the gradient, and the gradient is equal to 2.

Therefore,

10x - 3 = 2

10x = 2 + 3

10x = 5

x = $\frac{5}{10}$

x = $\frac{1}{2}$

Substitute the value of x in the equation,

y = 5x² - 3x + 1

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

y = 5$(\frac{1}{4} )$ - $(\frac{3}{2})$ + 1

y = $\frac{5}{4}$ - $\frac{3}{2}$ + 1

y = $\frac{5 - 6 + 4}{4}$

y = $\frac{3}{4}$

Final answer:

The co-ordinates of the point on graph y = 5x² - 3x + 1 where the gradient is 2 is $(\frac{1}{2},\frac{3}{4} )$ .