Equation for the Tangent to a Curve – Examples with answers

To find the equation of the tangent line, we can use the form $latex y=mx+b$, where m is the slope and b is the y-intercept.

The slope of the tangent line is found by evaluating the derivative of the function at the given point. That is, the slope is equal to $latex f'(1)$. Therefore, we have:

$latex f(x)=x^2$

$latex f'(x)=2x$

$latex m=f'(1)=2(1)$

$latex m=2$

So far, we have the equation $latex y=2x+b$. Now, we can find the value of b using the point (1, 3) in the equation:

$latex y=2x+b$

$latex 3=2(1)+b$

$latex b=1$

The equation of the tangent line at the point (1, 3) is $latex y=2x+1$.

We need the derivative of the function to find the slope of the tangent line. Thus, we have:

$latex f(x)=x^3-10x$

$latex f'(x)=3x^2-10$

To find the slope of the tangent line at the point (2, 1), we evaluate $latex f'(2)$:

$latex m=f'(2)=3(2)^2-10$

$latex m=2$

We form the equation $latex y=2x+b$. Now, we find the value of b, using the point (2, 1) in the equation:

$latex y=2x+b$

$latex 1=2(1)+b$

$latex b=-1$

The equation of the tangent line at the point (2, 1) is $latex y=2x-1$.

The derivative of the function allows us to find the slope of the tangent line. The derivative of the given function is:

$latex f(x)=2x^3-7x^2$

$latex f'(x)=6x^2-14x$

The slope of the tangent line at the point (2, 3) is equal to $latex f'(2)$. Thus, we have:

$latex m=f'(2)=6(2)^2-14(2)$

$latex m=24-28$

$latex m=-4$

Using the slope found, we have the equation $latex y=-4x+b$. Now, we find the value of b, using the point (2, 3) in the equation:

$latex y=-4x+b$

$latex 3=-4(2)+b$

$latex b=11$

The equation of the tangent line at the point (2, 3) is $latex y=-4x+11$.

We start by finding the derivative of the function:

$latex f(x)=x^3+8x^{-1}$

$latex f'(x)=3x^2-8x^{-2}$

$latex f'(x)=3x^2-\frac{8}{x^2}$

The slope of the tangent at the point (2, 4) is found by evaluating $latex f'(2)$. Therefore, we have:

$latex m=f'(2)=3(2)^2-\frac{8}{2^2}$

$latex =12-2$

$latex m=10$

We have the equation $latex y=10x+b$. Now, we use the point (2, 4) to find the value of b:

$latex y=10x+b$

$latex 4=10(2)+b$

$latex b=-15$

The equation of the tangent line at the point (2, 4) is $latex y=10x-15$.

We start by finding the derivative of the function to determine the slope of the tangent line:

$latex f(x)=-x^{-2}+x^{\frac{1}{2}}$

$$f'(x)=2x^{-3}+\frac{1}{2}x^{-\frac{1}{2}}$$

$$f'(x)=\frac{2}{x^3}+\frac{1}{2\sqrt{x}}$$

The slope of the tangent at the point (1, 3) is equal to $latex f'(1)$, so we have:

$$m=f'(1)=\frac{2}{(1)^3}+\frac{1}{2\sqrt{1}}$$

$latex =2+\frac{1}{2}$

$latex m=\frac{5}{2}$

Now that we have formed the equation $latex y=\frac{5}{2}x+b$, we use the point (1, 3) to find the value of b:

$$y=\frac{5}{2}x+b$$

$$3=\frac{5}{2}(1)+b$$

$latex b=\frac{1}{2}$

The equation of the tangent line at the point (1, 3) is $latex y=\frac{5}{2}x+\frac{1}{2}$.

We start by finding the derivative of the function:

$latex f(x)=\sin(x)-\cos(x)$

$latex f'(x)=\cos(x)+\sin(x)$

Now, we know that the slope of the tangent line at the point (0, 1) is found by evaluating $latex f'(0)$. Thus, we have:

$latex m=f'(0)=\cos(0)+\sin(0)$

$latex m=1+0$

$latex m=1$

So far, we have the equation $latex y=x+b$ and we can use this equation with the point (0, 1) to find the value of b, using the point (0, 1) in the equation:

$latex y=x+b$

$latex 1=0+b$

$latex b=1$

The equation of the tangent line at the point (0, 1) is $latex y=x+1$.

We need the derivative of the function to find the slope of the tangent. Thus, we have:

$latex f(x)=-2\sin(2x)+\cos(3x)$

$latex f'(x)=-4\cos(2x)-3\sin(3x)$

Now, we can evaluate $latex f'(0)$ to determine the slope of the tangent line at the point (0, 1):

$latex m=f'(0)=-4\cos(2(0))-3\sin(3(0))$

$latex m=-4+0$

$latex m=-4$

Using the slope found, we have the equation $latex y=-4x+b$. Using this equation with the point (0, 1) we find the value of b:

$latex y=-4x+b$

$latex 1=0+b$

$latex b=1$

The equation of the tangent line at the point (0, 1) is $latex y=-4x+1$.

The derivative of the function is:

$latex f(x)=x^2-3x+1$

$latex f'(x)=2x-3$

The statement tells us that we have to find the tangent line at the point where the curve intersects the y-axis. This happens when the coordinates in x are equal to 0. Then, we have:

$latex y=x^2-3x+1$

$latex y=0^2-3(0)+1$

$latex y=1$

Therefore, the point of the tangent is (0, 1). Now, we evaluate $latex f'(0)$ to find the slope of the tangent, and we have:

$latex m=f'(0)=2(0)-3$

$latex m=-3$

The slope found gives us the equation $latex y=-3x+b$. To find the value of b, we use the point (0, 1) in the equation:

$latex y=-3x+b$

$latex 1=-3(0)+b$

$latex b=1$

The equation of the tangent line at the point (0, 1) is $latex y=-3x+1$.

The points where the function intersects the x-axis are the zeros of the function. Therefore, we start by finding the zeros:

$latex x^2-5x+4=0$

$latex (x-4)(x-1)=0$

The function cuts the x-axis when $latex x=4$ and $latex x=1$. Also, the y-values at those points are 0, so we have (4, 0) and (1, 0).

Now, we find the derivative and evaluate $latex f'(4)$ and $latex f'(1)$ to determine the slopes of the tangent lines.

$latex f(x)=x^2-5x+4$

$latex f'(x)=2x-5$

$latex f'(4)=2(4)-5=3$

$latex f'(1)=2(1)-5=-3$

Now, we find the values of b using the form $latex y=mx+b$ with the slopes found and the coordinates of each point.

$latex 0=3(4)+b_{1}$

$latex b_{1}=-12$

$latex 0=-3(1)+b_{2}$

$latex b_{2}=3$

Therefore, the equations of the tangent lines are $latex y=3x-12$ and $latex y=-3x+3$.

The statement tells us that the lines are tangent at the points where $latex y=9$. Therefore, we have to find the values of x when $latex y=9$:

$latex x^2=9$

$latex x=\pm \sqrt{9}$

$latex x=3~~$ or $latex ~~x=-3$

Now, we use the derivative of the function to find the slopes of the lines by evaluating $latex f'(3)$ and $latex f'(-3)$.

$latex f(x)=x^2$

$latex f'(x)=2x$

$latex f'(3)=2(3)=6$

$latex f'(-3)=2(-3)=-6$

Using the form $latex y=mx+b$ with the slopes found and the coordinates of each point (the y-values are 9 in both cases), we can find the values of b:

$latex 9=6(3)+b_{1}$

$latex b_{1}=-9$

$latex 9=-6(-3)+b_{2}$

$latex b_{2}=-9$

Therefore, the equations of the tangent lines are $latex y=6x-9$ and $latex y=-6x-9$.