How do you find the equation of the tangent line to the graph of #f(x)=x^2+3x-7# at x =1?

Question

How do you find the equation of the tangent line to the graph of #f(x)=x^2+3x-7#  at x =1?

Luke Alvarez · Accepted Answer

y = 5x – 4

Remember this-

To find the equation of a straight line, we need two information –

They are (i) slope of the line and (ii) (x, y) co-ordinates of a point on that line.

Tangent is a straight line.

The given function is
y =# x^2 + 3x – 7#
It is a U shaped curve or parabola. It doesn’t have uniform slope throughout its length.

Its slope at any given point is its first derivative.

#dy/dx #= 2x + 3

At x = 1, its slope is # dy/dx# = 2(1) + 3 = 5

At x = 1 , the slope of the curve is 5.

A tangent is drawn to that point. To find the y co-ordinate of the point substitute x = 1 in the given function.

Y = 12 + 3(1) – 7 = -3

(1, -3) is a point on the tangent. The slope of the tangent is m = 5.
The equation of the tangent is

y – y1 = m(x – x1)
y – (- 3) = 5(x – 1)
y +3 = 5x – 1
y = 5x – 1 – 3

y = 5x – 4

Christopher Agresta · Answer

undefined To find the equation of the tangent line to the graph of f(x)=x^2+3x-7 at x=1, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the power rule: f'(x) = 2x + 3.

Next, we substitute x=1 into the derivative to find the slope of the tangent line: f'(1) = 2(1) + 3 = 5.

Now, we find the y-coordinate of the point on the graph of f(x) at x=1 by substituting x=1 into the original function: f(1) = (1)^2 + 3(1) - 7 = -3.

Therefore, the point on the tangent line is (1, -3) and the slope of the tangent line is 5.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as y - (-3) = 5(x - 1).

Simplifying, we get the equation of the tangent line as y = 5x - 8.