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Find the equation of tangent line to y=√2x+3 at x=3. Express in slope-intercept…
Related Topics
Wize University Calculus 1 Textbook > Derivatives
Tangent Lines
3 Activities
Find the equation of tangent line to
y
=
2
x
+
3
y=\sqrt{2x+3}
y
=
2
x
+
3
at
x
=
3
x=3
x
=
3
. Express in slope-intercept form.
Answer
I don't know
Check Submission
More Tangent Lines Questions:
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.
Tangent Lines
Find the point
(
a
,
b
)
\left(a,b\right)
(
a
,
b
)
at which the tangent line to the curve
y
=
2
x
2
+
3
x
−
7
y=2x^2+3x-7
y
=
2
x
2
+
3
x
−
7
is parallel to the tangent line of the curve
y
=
3
x
4
−
5
x
+
1
y=3x^4-5x+1
y
=
3
x
4
−
5
x
+
1
at the point
(
1
,
−
6
)
\left(1,-6\right)
(
1
,
−
6
)
.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Tangent Lines
Is there any point on the graph of
y
=
x
2
+
3
x
+
8
y = x^2+ 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
7
x
+
104
y = 7x + 104
y
=
7
x
+
104
?
Tangent Lines
The tangent line to the curve
y
=
−
x
2
−
3
y=-x^2-3
y
=
−
x
2
−
3
at
x
=
a
x=a
x
=
a
passes through
(
0
,
1
)
(0, 1)
(
0
,
1
)
. Find
a
a
a
and the corresponding function value.
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Practice: Tangent Through Outside Point
Q:
\textbf{Q:}
Q:
There are points on the curve
y
=
2
−
x
4
+
x
y=\dfrac{2-x}{4+x}
y
=
4
+
x
2
−
x
at which the tangent line passes through the origin. Find all such points.
Tangent Line Intercept
Let
f
(
x
)
=
3
x
2
−
4
x
+
9
f(x)=3x^2-4x+9
f
(
x
)
=
3
x
2
−
4
x
+
9
and let L denote the tangent line to the graph of
f
(
x
)
f(x)
f
(
x
)
at the point on the graph where
x
=
2
x=2
x
=
2
. Find the point where L intersects the x-axis.
Tangent Lines
Consider the line
y
=
4
x
+
3
y = 4x + 3
y
=
4
x
+
3
. To which of the following functions is it tangent at
x
=
1.
x = 1.
x
=
1.
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Consider the function
f
(
x
)
=
1
x
2
f(x) = \frac{1}{x^2}
f
(
x
)
=
x
2
1
. Let
g
(
a
)
g(a)
g
(
a
)
be the
x
x
x
- intercept of the tangent line of
f
(
x
)
f(x)
f
(
x
)
at
x
=
a
x = a
x
=
a
. Find
g
(
a
)
g(a)
g
(
a
)
for all
a
≠
0
a \not = 0
a
=
0
.
Given the function
f
(
x
)
=
x
3
−
3
x
+
1
f(x) = x^3 - 3x + 1
f
(
x
)
=
x
3
−
3
x
+
1
, find the equation of the tangent line at
x
=
2
x = 2
x
=
2
and find all other points that have tangent lines that are parallel to the one at
x
=
2
x = 2
x
=
2
.
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.
Finding Tangent and Normal Lines
Find the tangent line and normal line to
f
(
x
)
=
2
x
2
+
x
−
1
f(x)=2x^2+x-1
f
(
x
)
=
2
x
2
+
x
−
1
at
x
=
1
x=1
x
=
1
.
Tangent Lines
Let
f
(
x
)
=
1
3
x
3
+
1
2
x
2
+
5
x
−
4
f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2+5x-4
f
(
x
)
=
3
1
x
3
+
2
1
x
2
+
5
x
−
4
, and suppose
g(x)
is its inverse function. Find the
x
-coordinates of all points on the graph of
g(x)
so that the tangent line at those points are perpendicular to the function
y
=
−
7
x
+
2.
y=-7x+2.
y
=
−
7
x
+
2.
Tangent Lines
Find a horizontal tangent line of the curve
f
(
x
)
=
x
3
/
3
−
2
x
2
−
5
x
+
4
f(x)=x^3/3-2x^2-5x+4
f
(
x
)
=
x
3
/3
−
2
x
2
−
5
x
+
4
Tangent Lines
Find all points on the graph of
f
(
x
)
=
2
x
2
+
x
−
3
f(x)=2x^2+x-3
f
(
x
)
=
2
x
2
+
x
−
3
whose tangent line passes through the point (3,0).
In the function
C
=
−
4
k
t
+
3
C=-4kt+3
C
=
−
4
k
t
+
3
, where
C
C
C
represents the concentration of a certain substance in a tank of water,
t
t
t
represents time, and
k
k
k
is a positive constant, the rate of change of the dependent variable with respect to the independent variable is a constant. The statement is:
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Tangent Lines
Find the horizontal tangent line of the curve
f
(
x
)
=
x
2
3
−
2
x
2
−
5
x
+
4
\displaystyle f(x)=\frac{x^2}{3}-2x^2-5x+4
f
(
x
)
=
3
x
2
−
2
x
2
−
5
x
+
4
Tangent Lines
Is there any point on the graph of
y
=
x
2
+
3
x
+
8
y = x^2+ 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
7
x
+
104
y = 7x + 104
y
=
7
x
+
104
?
Tangent Lines
The tangent line to the curve
y
=
−
x
2
−
3
y=-x^2-3
y
=
−
x
2
−
3
at
x
=
a
x=a
x
=
a
passes through
(
0
,
1
)
(0, 1)
(
0
,
1
)
. Find
a
a
a
and the corresponding function value.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Practice: Tangent Through Outside Point
Q:
\textbf{Q:}
Q:
There are points on the curve
y
=
2
−
x
4
+
x
y=\dfrac{2-x}{4+x}
y
=
4
+
x
2
−
x
at which the tangent line passes through the origin. Find all such points.
Tangent Line Intercept
Let
f
(
x
)
=
3
x
2
−
4
x
+
9
f(x)=3x^2-4x+9
f
(
x
)
=
3
x
2
−
4
x
+
9
and let L denote the tangent line to the graph of
f
(
x
)
f(x)
f
(
x
)
at the point on the graph where
x
=
2
x=2
x
=
2
. Find the point where L intersects the x-axis.
Find the point on the graph
y
2
= 4
x
+ 5 where the tangent line is parallel to the line
y
= 2
x
+ 184.
Tangent Lines
At which point on the graph
y
=
2
x
+
5
y=\sqrt{2x+5}
y
=
2
x
+
5
the tangent line is parallel to the line with the equation
y
=
1
3
x
+
11
?
y=\frac{1}{3}x+11 ?
y
=
3
1
x
+
11
?
If
f
(
4
)
=
1
f(4)=1
f
(
4
)
=
1
,
f
′
(
4
)
=
−
1
f'(4)=-1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x)=\frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
How many distinct horizontal lines are tangent to the equation
f
(
x
)
=
x
2
(
9
−
x
2
)
f\left(x\right)=x^2\left(9-x^2\right)
f
(
x
)
=
x
2
(
9
−
x
2
)
?
🌶️
TOUGH!
As shown in the figure below, the tangent line to the graph
y
=
f
(
x
) intersects the
x
-axis at
x
=
b
. What is the value of
b
in terms of
a
,
f
(
a
), and
f
’(
a
)?
Find the point where the tangent line is horizontal for the curve
f
(
x
)
=
x
2
3
−
2
x
2
−
5
x
+
4
f(x)=\frac{x^2}{3}-2x^2-5x+4
f
(
x
)
=
3
x
2
−
2
x
2
−
5
x
+
4
.
Find the
y
y
y
-intercept of the tangent line of
f
(
x
)
=
1
x
+
x
at
x
=
2
f(x)=\frac{1}{\sqrt{x}}+\sqrt{x} \ \ \text{at }\ x=2
f
(
x
)
=
x
1
+
x
at
x
=
2
.
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
🌶️
TOUGH!
Let
f
(
x
)
=
x
2
f(x)=x^2
f
(
x
)
=
x
2
, given the tangent line of
f
f
f
at
x
=
a
x=a
x
=
a
passes through the point
(
4
,
15
)
(4,15)
(
4
,
15
)
not on the function, find
a
a
a
.
Find the equation of the tangent line to
f
(
x
)
=
(
x
2
−
2
)
sin
x
+
2
x
cos
x
f(x) = (x^2 - 2) \sin x+2x \cos x
f
(
x
)
=
(
x
2
−
2
)
sin
x
+
2
x
cos
x
at
x
=
π
x = \pi
x
=
π
.
Practice: Equation of tangent (and normal) line
Find the tangent and normal lines to the graph
f
(
x
)
=
x
f\left(x\right)=\sqrt{x}
f
(
x
)
=
x
at
x = 9
. Which point on this graph is the slope of the tangent line parallel to the line
y
=
3
x
+
1
?
y = 3x + 1?
y
=
3
x
+
1
?
Find the equation of the tangent line to the given point for the following function.
f
(
x
)
=
e
x
2
+
x
,
at
P
(
1
,
e
2
)
f(x)=e^{x^2+\sqrt{x}},\,\text{at}\,P(1,e^2)
f
(
x
)
=
e
x
2
+
x
,
at
P
(
1
,
e
2
)
Tangent Lines
Find the point
(
a
,
b
)
\left(a,b\right)
(
a
,
b
)
at which the tangent line to the curve
y
=
2
x
2
+
3
x
−
7
y=2x^2+3x-7
y
=
2
x
2
+
3
x
−
7
is parallel to the tangent line of the curve
y
=
3
x
4
−
5
x
+
1
y=3x^4-5x+1
y
=
3
x
4
−
5
x
+
1
at the point
(
1
,
−
6
)
\left(1,-6\right)
(
1
,
−
6
)
.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
?
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Tangent Lines
Consider the function
f
(
x
)
=
−
2
x
2
+
3
x
−
1
f(x) = - 2x^2 + 3x - 1
f
(
x
)
=
−
2
x
2
+
3
x
−
1
. There is a point
(
a
,
b
)
(a, b)
(
a
,
b
)
such that the tangent line to
f
f
f
at
(
a
,
b
)
(a, b)
(
a
,
b
)
is perpendicular to the line
y
=
x
3
+
5
y = \frac{x}{3} + 5
y
=
3
x
+
5
. Find the point
(
a
,
b
)
(a, b)
(
a
,
b
)
and write down the equation of the tangent line.
Evaluate the slope of the tangent line to the curve at the given point.
x
3
+
y
3
=
4
x
y
,
at
P
(
2
,
2
)
x^3+y^3=4xy ,\,\, \text{at} \,\, P(2,2)
x
3
+
y
3
=
4
x
y
,
at
P
(
2
,
2
)