Calculus Examples
y=x2+3x+34y=x2+3x+34 , (0,34)(0,34)
Step 1
Write y=x2+3x+34y=x2+3x+34 as a function.
f(x)=x2+3x+34f(x)=x2+3x+34
Step 2
Step 2.1
Evaluate f(x)=x2+3x+34f(x)=x2+3x+34 at x=0x=0.
Step 2.1.1
Replace the variable xx with 00 in the expression.
f(0)=(0)2+3(0)+34f(0)=(0)2+3(0)+34
Step 2.1.2
Simplify the result.
Step 2.1.2.1
Simplify each term.
Step 2.1.2.1.1
Raising 00 to any positive power yields 00.
f(0)=0+3(0)+34f(0)=0+3(0)+34
Step 2.1.2.1.2
Multiply 33 by 00.
f(0)=0+0+34f(0)=0+0+34
f(0)=0+0+34f(0)=0+0+34
Step 2.1.2.2
Simplify by adding numbers.
Step 2.1.2.2.1
Add 00 and 00.
f(0)=0+34f(0)=0+34
Step 2.1.2.2.2
Add 00 and 3434.
f(0)=34f(0)=34
f(0)=34f(0)=34
Step 2.1.2.3
The final answer is 3434.
3434
3434
3434
Step 2.2
Since 34=3434=34, the point is on the graph.
The point is on the graph
The point is on the graph
Step 3
The slope of the tangent line is the derivative of the expression.
mm == The derivative of f(x)=x2+3x+34f(x)=x2+3x+34
Step 4
Consider the limit definition of the derivative.
f′(x)=limh→0f(x+h)-f(x)h
Step 5
Step 5.1
Evaluate the function at x=x+h.
Step 5.1.1
Replace the variable x with x+h in the expression.
f(x+h)=(x+h)2+3(x+h)+34
Step 5.1.2
Simplify the result.
Step 5.1.2.1
Simplify each term.
Step 5.1.2.1.1
Rewrite (x+h)2 as (x+h)(x+h).
f(x+h)=(x+h)(x+h)+3(x+h)+34
Step 5.1.2.1.2
Expand (x+h)(x+h) using the FOIL Method.
Step 5.1.2.1.2.1
Apply the distributive property.
f(x+h)=x(x+h)+h(x+h)+3(x+h)+34
Step 5.1.2.1.2.2
Apply the distributive property.
f(x+h)=x⋅x+xh+h(x+h)+3(x+h)+34
Step 5.1.2.1.2.3
Apply the distributive property.
f(x+h)=x⋅x+xh+hx+h⋅h+3(x+h)+34
f(x+h)=x⋅x+xh+hx+h⋅h+3(x+h)+34
Step 5.1.2.1.3
Simplify and combine like terms.
Step 5.1.2.1.3.1
Simplify each term.
Step 5.1.2.1.3.1.1
Multiply x by x.
f(x+h)=x2+xh+hx+h⋅h+3(x+h)+34
Step 5.1.2.1.3.1.2
Multiply h by h.
f(x+h)=x2+xh+hx+h2+3(x+h)+34
f(x+h)=x2+xh+hx+h2+3(x+h)+34
Step 5.1.2.1.3.2
Add xh and hx.
Step 5.1.2.1.3.2.1
Reorder x and h.
f(x+h)=x2+hx+hx+h2+3(x+h)+34
Step 5.1.2.1.3.2.2
Add hx and hx.
f(x+h)=x2+2hx+h2+3(x+h)+34
f(x+h)=x2+2hx+h2+3(x+h)+34
f(x+h)=x2+2hx+h2+3(x+h)+34
Step 5.1.2.1.4
Apply the distributive property.
f(x+h)=x2+2hx+h2+3x+3h+34
f(x+h)=x2+2hx+h2+3x+3h+34
Step 5.1.2.2
The final answer is x2+2hx+h2+3x+3h+34.
x2+2hx+h2+3x+3h+34
x2+2hx+h2+3x+3h+34
x2+2hx+h2+3x+3h+34
Step 5.2
Reorder.
Step 5.2.1
Move 3x.
x2+2hx+h2+3h+3x+34
Step 5.2.2
Move x2.
2hx+h2+x2+3h+3x+34
Step 5.2.3
Reorder 2hx and h2.
h2+2hx+x2+3h+3x+34
h2+2hx+x2+3h+3x+34
Step 5.3
Find the components of the definition.
f(x+h)=h2+2hx+x2+3h+3x+34
f(x)=x2+3x+34
f(x+h)=h2+2hx+x2+3h+3x+34
f(x)=x2+3x+34
Step 6
Plug in the components.
f′(x)=limh→0h2+2hx+x2+3h+3x+34-(x2+3x+34)h
Step 7
Step 7.1
Simplify the numerator.
Step 7.1.1
Apply the distributive property.
f′(x)=limh→0h2+2hx+x2+3h+3x+34-x2-(3x)-1⋅34h
Step 7.1.2
Simplify.
Step 7.1.2.1
Multiply 3 by -1.
f′(x)=limh→0h2+2hx+x2+3h+3x+34-x2-3x-1⋅34h
Step 7.1.2.2
Multiply -1 by 34.
f′(x)=limh→0h2+2hx+x2+3h+3x+34-x2-3x-34h
f′(x)=limh→0h2+2hx+x2+3h+3x+34-x2-3x-34h
Step 7.1.3
Subtract x2 from x2.
f′(x)=limh→0h2+2hx+3h+3x+34+0-3x-34h
Step 7.1.4
Add h2 and 0.
f′(x)=limh→0h2+2hx+3h+3x+34-3x-34h
Step 7.1.5
Subtract 3x from 3x.
f′(x)=limh→0h2+2hx+3h+0+34-34h
Step 7.1.6
Add h2 and 0.
f′(x)=limh→0h2+2hx+3h+34-34h
Step 7.1.7
Subtract 34 from 34.
f′(x)=limh→0h2+2hx+3h+0h
Step 7.1.8
Add h2+2hx+3h and 0.
f′(x)=limh→0h2+2hx+3hh
Step 7.1.9
Factor h out of h2+2hx+3h.
Step 7.1.9.1
Factor h out of h2.
f′(x)=limh→0h⋅h+2hx+3hh
Step 7.1.9.2
Factor h out of 2hx.
f′(x)=limh→0h(h)+h(2x)+3hh
Step 7.1.9.3
Factor h out of 3h.
f′(x)=limh→0h(h)+h(2x)+h⋅3h
Step 7.1.9.4
Factor h out of h(h)+h(2x).
f′(x)=limh→0h(h+2x)+h⋅3h
Step 7.1.9.5
Factor h out of h(h+2x)+h⋅3.
f′(x)=limh→0h(h+2x+3)h
f′(x)=limh→0h(h+2x+3)h
f′(x)=limh→0h(h+2x+3)h
Step 7.2
Reduce the expression by cancelling the common factors.
Step 7.2.1
Cancel the common factor of h.
Step 7.2.1.1
Cancel the common factor.
f′(x)=limh→0h(h+2x+3)h
Step 7.2.1.2
Divide h+2x+3 by 1.
f′(x)=limh→0h+2x+3
f′(x)=limh→0h+2x+3
Step 7.2.2
Reorder h and 2x.
f′(x)=limh→02x+h+3
f′(x)=limh→02x+h+3
f′(x)=limh→02x+h+3
Step 8
Step 8.1
Split the limit using the Sum of Limits Rule on the limit as h approaches 0.
limh→02x+limh→0h+limh→03
Step 8.2
Evaluate the limit of 2x which is constant as h approaches 0.
2x+limh→0h+limh→03
Step 8.3
Evaluate the limit of 3 which is constant as h approaches 0.
2x+limh→0h+3
2x+limh→0h+3
Step 9
Evaluate the limit of h by plugging in 0 for h.
2x+0+3
Step 10
Add 2x and 0.
2x+3
Step 11
Step 11.1
Multiply 2 by 0.
m=0+3
Step 11.2
Add 0 and 3.
m=3
m=3
Step 12
The slope is m=3 and the point is (0,34).
m=3,(0,34)
Step 13
Step 13.1
Use the formula for the equation of a line to find b.
y=mx+b
Step 13.2
Substitute the value of m into the equation.
y=(3)⋅x+b
Step 13.3
Substitute the value of x into the equation.
y=(3)⋅(0)+b
Step 13.4
Substitute the value of y into the equation.
34=(3)⋅(0)+b
Step 13.5
Find the value of b.
Step 13.5.1
Rewrite the equation as (3)⋅(0)+b=34.
(3)⋅(0)+b=34
Step 13.5.2
Simplify (3)⋅(0)+b.
Step 13.5.2.1
Multiply 3 by 0.
0+b=34
Step 13.5.2.2
Add 0 and b.
b=34
b=34
b=34
b=34
Step 14
Now that the values of m (slope) and b (y-intercept) are known, substitute them into y=mx+b to find the equation of the line.
y=3x+34
Step 15