# If the tangent line to y = f(x) at (4,3) passes through the point (0,2), Find f(4) and f'(4)? An explanation would also be very helpful.

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Mar 4, 2017

$f \left(4\right) = 3$

$f ' \left(4\right) = \frac{1}{4}$

#### Explanation:

The question gives you $f \left(4\right)$ already, because the point $\left(4 , 3\right)$ is given. When $x$ is $4$, $\left[y = f \left(x\right) =\right] f \left(4\right)$ is $3$.

We can find $f ' \left(4\right)$ by finding the gradient at the point $f \left(4\right)$, which we can do because we know the tangent touches both $\left(4 , 3\right)$ and $\left(0 , 2\right)$.

The gradient of a line is given by rise over run, or the change in $y$ divided by the change in $x$, or, mathematically,

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

We know two points on the graph in the question, so effectively we know the two values we need for $y$ and $x$ each. Say that

$\left(0 , 2\right) \to {x}_{1} = 0 , {y}_{1} = 2$

$\left(4 , 3\right) \to {x}_{2} = 4 , {y}_{2} = 3$

so

$m = \frac{3 - 2}{4 - 0} = \frac{1}{4}$

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