Given the polynomial 5x3−2x2+12, what is the leading coefficient?
1 answer:
You might be interested in
Answer:
y = 13*( -x/9 + 1/5)
Step-by-step explanation:
Given:
- The curve has an equation as follows:
Find:
a. Verify that the given point (2,2) lies on the curve.
b. Determine an equation of the line tangent to the curve at the given point.
Solution:
- To verify whether the point lies on the given curve we will substitute the coordinates of the point into the equation as follows:
44 = 5*(2)^2 + 3*(2)(2) + 3*(2)^2
44 = 20 + 12 + 12
44 = 44 ......Hence proven.
- The equation of the line tangent to the curve is expressed as a linear function as follows:
y = m*x + C
Where, m is the gradient of the line.
C is the y-intercept.
m = Δy / Δx = dy/dx
- We will take the derivative of the given curve with respect to x as follows:
- Evaluate y' at the point (2,2) we get:
y' = - ( 10(2) + 3(2) ) / ( 3(2) + 6(2) )
y' = - ( 26 ) / (18)
y'= m = - 13/9
- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:
y = -13*x/9 + C
2 =-13*(2)/9 + C
C = 13 / 5
- The equation of the tangent is as follows:
y = 13*( -x/9 + 1/5)