Find the equation of the tangent to a curve y= 3x² - 6x + 1 at a point

y

=

3

x

−

1

Explanation:

Find  

d

y

d

x

of  

y

equation

y

=

3

x

2

−

x

3

d

y

d

x

=

6

x

−

3

x

2

Then from the point (1,2) we can obtain the x value which is 1

Insert  

x

=

1

into  

d

y

d

x

d

y

d

x

=

6

(

1

)

−

3

(

1

)

2

d

y

d

x

=

3

To form the equation we have to base it from the original equation

y

=

m

x

+

c

we know that m=3 from  

d

y

d

x

=

3

y

=

3

x

+

c

use the coordinates given fro x and y values to find c

(1,2) x=1 , y=2

(

2

)

=

3

(

1

)

+

c

c

=

−

1

Then we put c back into the general form to get the final tangent equation

y

=

3

x

−

1

Step-by-step explanation:

A standard normal distribution refers to a normal distribution with a mean of 0 and a standard deviation of 1. To solve this proble we're going to need the help of a calculator:

(a) P(0 ≤ Z ≤ 2.38) = 0.4913

(b) P(0 ≤ Z ≤ 1) = 0.3413

(c) P(−2.70 ≤ Z ≤ 0) = 0.4965

(d) P(−2.70 ≤ Z ≤ 2.70) = 0.9931

(e) P(Z ≤ 1.62) = 0.9474

(f) P(−1.55 ≤ Z)= 0.9394

(g) P(−1.70 ≤ Z ≤ 2.00) = 0.9327

(h) P(1.62 ≤ Z ≤ 2.50) = 0.0464

(i) P(1.70 ≤ Z) = 0.0445

(j) P(|Z| ≤ 2.50) = 0.9876

All values are verified! ✅