If f(x)=x62+5x-6 what is the equation of the tangent to the curve when the gradient of the tangent is 11?

Question

I really need help please.

step by step points!

Answer 1

Hi,

For x² + 5x - 6, the first derivative is f'(x) = 2x + 5. If this equals 11, then 2x + 5 = 11.
2x = 6
x = 3.
This occurs on the graph of f(x) = x² + 5x - 6 when x = 3. that is actually at the point (3,18) from substituting 3 for x in f(x).

So the line through (3,18) with a gradient of 11 is:

y - y1 = m(x - x1) <== point slope form
y - 18 = 11(x - 3)
y - 18 = 11x - 33
y = 11x - 15

So y = 11x - 15 is the equation of the tangent line in slope-intercept form.

I hope that helps!! :-)

Answer 2

f(x) = x^2 + 5x -6
Gradient is equal to derivative of f(x).
f'(x) = 2x + 5
Subst. gradient = 11
11= 2x + 5
x = (11-5)/2 = 3
Therefore the point of intersection between this tangent and the curve is at x = 3.
Subst. x = 3 into original equation.
f(x) = (3) ^ 2 + 5(3) - 6 = 18
Therefore coordination of intersection is (3, 18)
Subst. this and the gradient = 11 into the point-gradient formula.
y-18 = 11(x-3)
y-18 = 11x - 33
Therefore the equation of the tangent to the curve is:
11x - y -15 = 0

Answer 3

I believe you meant f(x)=x^2+5x-6
The gradient of the tangent is the same as the gradient of the function at any given point.

The gradient is given by differentiation. When the function is in the form f(x)=ax^n, then the gradient function, f'(x)=nax^(n-1).

And to differentiate a constant value, with no "x", it is zero, since this constant does not affect the gradient of the graph.

Also, when there are more than one terms, just add them together to find the derivative. This is differentiation by rule.

So the gradient funtion of this one, f(x)=x^2+5x-6
Differentiating y=x^2 gives f'(x)=2x
differentiating y=5x gives f'(x)=5
differentiating y=-6 gives f'(x)=0.
put them together,

therefore, f'(x)=2x+5

So the gradient of the tangent is 11. Therefore, f'(x)=11, since the gradient of the tangent is that of the function at that point.

11=2x+5
6=2x
x=3

so grad is 11 when x=3.

Substitute x=3 into f(x), to find y-value of that function at the tangent.
f(3)=(3)^2+5(3)-6
=9+15-6
=18
so this occurs at (3,18)

The tangent of the curve occurs at (3,18), and has a gradient of 11. So:

y-y1=m(x-x1)
y-18=11(x-3)
y=11x-33+18
y=11x-15

Answer 4

f(x) = x^2 + 5x -- 6
f'(x) = 11 => 2x + 5 = 11 => x = 3
y = f(3) = 3^2 + 5*3 -- 6 = 18
the point (3, 18) on curve f(x) = x^2 + 5x -- 6 has tangent whose gradient is 11.
equation to that tangent is
y -- 18 = 11(x -- 3) or y = 11x - 15

Answer 5

Assuming f(x) = x² + 5x - 6
f `(x) = 2x + 5
2x + 5 = 11
2x = 6
x = 3
f(3) = 9 + 15 - 6
f(3) = 18
m = 11
Tangent passes thro` (3 , 18)
Equation of tangent is:-
y - 18 = 11 (x - 3)
y = 11x - 33 + 18
y = 11x - 15

Answer 6

are u sure the equation that u give is right? for graph to be curve, the highest power of x can't be 1