Equation of a line?

Question

Find the equation of the line tangent to the curve Y= x^2 + 4X + 2 at the point x=1.

Answer 1

Y(1) = 1+4+2 = 7
This tells you the tangent point is at (1,7)

Next take the derivative of Y(x) to find the slope at any given point
Y'(x) = 2x+4

Find the slope at x=1
Y'(1) = 2+4 = 6

Therefore the slope M=6 and passes through the point (1,7)
Y=mx+b = 6x+b
7=6*1+b
b = 1

Therefore
y = 6x+1 is a line that is tangent to Y= x^2 + 4X + 2 at x=1

Answer 2

The derivative of the curve is 2x+4, which is the slope of the curve at any given point x.
If x=1, then the slope of the tangent line is 6 (m=6).
The point on the curve at x=1 is the point (1,7), so we use the point-slope form to determine the tangent line.
y-(7)=6(x-1)
y-7=6x-6
so
y=6x+1

Answer 3

By finding the derivative of the curve equation, you will get the slope of the tangent by putting the value of x(=1). and you can use the formule y-y1=m(x-x1).
so, the answer is y-7=6(x-1).=> 6x-y+1=0 is the required equation fo the tanget to the curve.

Answer 4

Your first thought should be "I need to take the derivative."
y' = 2x + 4
Now you need to find the slope.
To find the slope, m, plug in the given x value.
m = 2(1) + 4
m = 6
Now you need to find the line.
Use the equation
y - y1 = m(x-x1), where m is the slope and (x1,y1) is the point.
x1 = 1 is given. To find y1, replace x with 1 and solve for y
y = (1)^2 + 4(1) + 2
y = 1+4+2
y = 7
Your equation is
y - 7 = 6(x-1)

Answer 5

f(x)=x^2+4x+2
f'(x)=2x+4
at x=1 f'(x)=6
at x=1 y=7
equation of the tangent
y-7=6(x-1)
y-7=6x-6
6x-y+1=0 is the equation of the tangent