Question

What is the equation of the line tangent to `f(x)=5x^2-3x+2` at `x=3`?

Answer 1

y = 27x -43

Explanation:

We can write the equation in slope-intercept form y = mx +c

where m represents the gradient and c , the y-intercept.

The value of f'(3) is equal to m and c may be found by evaluating f(3)

`f(x)=5x^2-3x+2`

`rArr f'(x)=10x-3`

and f'(3) = 10(3) - 3 = 27 = m (gradient of tangent)

Partial equation is therefore : y = 27x +c

now `f(3)=5(3)^2-3(3)+2=38`

hence (3 ,38) is a point on the tangent and substituting this into the partial equation will give us the value of c.

x = 3 , y = 38 : 38 = 81 + c → c = - 43

Thus the equation of tangent is `color(red)(|bar(ul(color(white)(a/a)color(black)(y=27x-43)color(white)(a/a)|)))`