Find an equation of the tangent to the curve (y=x^5+20x^2-9) at a the point where x=1

please explain the method

thank you

1. Start with your original equation then find the first derivative.

y = x^5 + 20x^2 - 9 becomes 5x^4 +40x

2. Now this new equation will give you the gradient of the tangent at any point of the curve. We want when x=1 so substitute 1 for x.

5(1^4) + 40(1) = 45 so gradient is 45.

3. Using the formula for a straight line y = mx + c where m is the gradiant you now know that the line is y = 45x + c

4. Substitute x = 1 into the original equation to find out the y value and the full coordinate.

1 + 20 - 9 = 12 so coordinate is (1,12)

5. Substitute this coordinate into the y = 45x + c to get 12 = 45 + c from this you know that c must be -33

6. Your full equation for the tangent of the curve at x=1 is y= 45x - 33!

I have checked this twice and even plotted this on a graph for you. I assure you the answer is y=45x-33 not y=45x+11


Hope I Helped.

If x = 1 then y = 1^5 + 20(1^2) - 9 = 1 + 20*1 - 9 = 1 + 20 - 9 = 12, so the tangent passes through (1, 12)

y' = 5x^4 + 40x
If x = 1 then y' = 5(1^4) + 40*1 = 5*1 + 40 = 5 + 40 = 45
So the slope of the tangent line is 45
So the tangent line is y = 45x + b, which passes through (1, 12), so we have:
12 = 45*1 + b
12 = 45 + b
b = 12 - 45
b = -33
So the tangent line is y = 45x - 33

y(1) = 1+20-9 = 12
y'(1) = 5+40 = 45
Answer: y - 12 = 45(x - 1)