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Question
x=y^{2}+2y+2
Identify the conic
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Find the standard equation of the parabola
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Find the vertex of the parabola
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Find the focus of the parabola
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Find the directrix of the parabola
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\left(y+1\right)^{2}=x-1
Evaluate
x=y^{2}+2y+2
Swap the sides of the equation
y^{2}+2y+2=x
Move the constant to the right-hand side and change its sign
y^{2}+2y=x-2
To complete the square, the same value needs to be added to both sides
y^{2}+2y+1=x-2+1
\text{Use }a^2+2ab+b^2=(a+b)^2\text{ to factor the expression}
\left(y+1\right)^{2}=x-2+1
Solution
\left(y+1\right)^{2}=x-1
Show Solutions
Solve the equation
\begin{align}&y=-1+\sqrt{-1+x}\\&y=-1-\sqrt{-1+x}\end{align}
Evaluate
x=y^{2}+2y+2
Swap the sides of the equation
y^{2}+2y+2=x
Move the expression to the left side
y^{2}+2y+2-x=0
\text{Substitute a=}1\text{,b=}2\text{ and c=}2-x\text{ into the quadratic formula }y\text{=}\frac{-b\pm\sqrt{b^2-4ac}}{2a}
y=\frac{-2\pm \sqrt{2^{2}-4\left(2-x\right)}}{2}
Simplify the expression
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Evaluate
2^{2}-4\left(2-x\right)
Multiply the terms
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Evaluate
4\left(2-x\right)
Apply the distributive property
4\times 2-4x
Multiply the numbers
8-4x
2^{2}-\left(8-4x\right)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
2^{2}-8+4x
Evaluate the power
4-8+4x
Subtract the numbers
-4+4x
y=\frac{-2\pm \sqrt{-4+4x}}{2}
Simplify the radical expression
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Evaluate
\sqrt{-4+4x}
Factor the expression
\sqrt{4\left(-1+x\right)}
The root of a product is equal to the product of the roots of each factor
\sqrt{4}\times \sqrt{-1+x}
Evaluate the root
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Evaluate
\sqrt{4}
\text{Write the number in exponential form with the base of }2
\sqrt{2^{2}}
\text{Reduce the index of the radical and exponent with }2
2
2\sqrt{-1+x}
y=\frac{-2\pm 2\sqrt{-1+x}}{2}
\text{Separate the equation into }2\text{ possible cases}
\begin{align}&y=\frac{-2+2\sqrt{-1+x}}{2}\\&y=\frac{-2-2\sqrt{-1+x}}{2}\end{align}
Simplify the expression
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Evaluate
y=\frac{-2+2\sqrt{-1+x}}{2}
Divide the terms
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Evaluate
\frac{-2+2\sqrt{-1+x}}{2}
Rewrite the expression
\frac{2\left(-1+\sqrt{-1+x}\right)}{2}
Reduce the fraction
-1+\sqrt{-1+x}
y=-1+\sqrt{-1+x}
\begin{align}&y=-1+\sqrt{-1+x}\\&y=\frac{-2-2\sqrt{-1+x}}{2}\end{align}
Solution
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Evaluate
y=\frac{-2-2\sqrt{-1+x}}{2}
Divide the terms
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Evaluate
\frac{-2-2\sqrt{-1+x}}{2}
Rewrite the expression
\frac{2\left(-1-\sqrt{-1+x}\right)}{2}
Reduce the fraction
-1-\sqrt{-1+x}
y=-1-\sqrt{-1+x}
\begin{align}&y=-1+\sqrt{-1+x}\\&y=-1-\sqrt{-1+x}\end{align}
Show Solutions
Testing for symmetry
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Testing for symmetry about the origin
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Testing for symmetry about the x-axis
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Testing for symmetry about the y-axis
\textrm{Not symmetry with respect to the origin}
Evaluate
x=y^{2}+2y+2
\text{To test if the graph of }x=y^{2}+2y+2\text{ is symmetry with respect to the origin,substitute -x for x and -y for y}
-x=\left(-y\right)^{2}+2\left(-y\right)+2
Evaluate
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Evaluate
\left(-y\right)^{2}+2\left(-y\right)+2
Multiply the numbers
\left(-y\right)^{2}-2y+2
Rewrite the expression
y^{2}-2y+2
-x=y^{2}-2y+2
Solution
\textrm{Not symmetry with respect to the origin}
Show Solutions
Find the first derivative
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\text{Find the derivative with respect to }x
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\text{Find the derivative with respect to }y
\frac{dy}{dx}=\frac{1}{2y+2}
Calculate
x=y^{2}+2y+2
Take the derivative of both sides
\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(y^{2}+2y+2\right)
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
1=\frac{d}{dx}\left(y^{2}+2y+2\right)
Calculate the derivative
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Evaluate
\frac{d}{dx}\left(y^{2}+2y+2\right)
Use differentiation rules
\frac{d}{dx}\left(y^{2}\right)+\frac{d}{dx}\left(2y\right)+\frac{d}{dx}\left(2\right)
Evaluate the derivative
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Evaluate
\frac{d}{dx}\left(y^{2}\right)
Use differentiation rules
\frac{d}{dy}\left(y^{2}\right)\times \frac{dy}{dx}
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
2y\frac{dy}{dx}
2y\frac{dy}{dx}+\frac{d}{dx}\left(2y\right)+\frac{d}{dx}\left(2\right)
Evaluate the derivative
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Evaluate
\frac{d}{dx}\left(2y\right)
Use differentiation rules
\frac{d}{dy}\left(2y\right)\times \frac{dy}{dx}
Evaluate the derivative
2\frac{dy}{dx}
2y\frac{dy}{dx}+2\frac{dy}{dx}+\frac{d}{dx}\left(2\right)
\text{Use }\frac{d}{dx}(c)=0\text{ to find derivative}
2y\frac{dy}{dx}+2\frac{dy}{dx}+0
Evaluate
2y\frac{dy}{dx}+2\frac{dy}{dx}
1=2y\frac{dy}{dx}+2\frac{dy}{dx}
Swap the sides of the equation
2y\frac{dy}{dx}+2\frac{dy}{dx}=1
Collect like terms by calculating the sum or difference of their coefficients
\left(2y+2\right)\frac{dy}{dx}=1
Divide both sides
\frac{\left(2y+2\right)\frac{dy}{dx}}{2y+2}=\frac{1}{2y+2}
Solution
\frac{dy}{dx}=\frac{1}{2y+2}
Show Solutions
Find the second derivative
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\text{Find the second derivative with respect to }x
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\text{Find the second derivative with respect to }y
\frac{d^2y}{dx^2}=-\frac{1}{4y^{3}+12y^{2}+12y+4}
Calculate
x=y^{2}+2y+2
Take the derivative of both sides
\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(y^{2}+2y+2\right)
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
1=\frac{d}{dx}\left(y^{2}+2y+2\right)
Calculate the derivative
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Evaluate
\frac{d}{dx}\left(y^{2}+2y+2\right)
Use differentiation rules
\frac{d}{dx}\left(y^{2}\right)+\frac{d}{dx}\left(2y\right)+\frac{d}{dx}\left(2\right)
Evaluate the derivative
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Evaluate
\frac{d}{dx}\left(y^{2}\right)
Use differentiation rules
\frac{d}{dy}\left(y^{2}\right)\times \frac{dy}{dx}
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
2y\frac{dy}{dx}
2y\frac{dy}{dx}+\frac{d}{dx}\left(2y\right)+\frac{d}{dx}\left(2\right)
Evaluate the derivative
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Evaluate
\frac{d}{dx}\left(2y\right)
Use differentiation rules
\frac{d}{dy}\left(2y\right)\times \frac{dy}{dx}
Evaluate the derivative
2\frac{dy}{dx}
2y\frac{dy}{dx}+2\frac{dy}{dx}+\frac{d}{dx}\left(2\right)
\text{Use }\frac{d}{dx}(c)=0\text{ to find derivative}
2y\frac{dy}{dx}+2\frac{dy}{dx}+0
Evaluate
2y\frac{dy}{dx}+2\frac{dy}{dx}
1=2y\frac{dy}{dx}+2\frac{dy}{dx}
Swap the sides of the equation
2y\frac{dy}{dx}+2\frac{dy}{dx}=1
Collect like terms by calculating the sum or difference of their coefficients
\left(2y+2\right)\frac{dy}{dx}=1
Divide both sides
\frac{\left(2y+2\right)\frac{dy}{dx}}{2y+2}=\frac{1}{2y+2}
Divide the numbers
\frac{dy}{dx}=\frac{1}{2y+2}
Take the derivative of both sides
\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1}{2y+2}\right)
Calculate the derivative
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{2y+2}\right)
Rewrite the expression in exponential form
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\left(2y+2\right)^{-1}\right)
Calculate the derivative
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Evaluate
\frac{d}{dx}\left(\left(2y+2\right)^{-1}\right)
Evaluate the derivative
-\left(2y+2\right)^{-2}\times \frac{d}{dx}\left(2y+2\right)
Evaluate the derivative
-\left(2y+2\right)^{-2}\times 2\frac{dy}{dx}
Calculate
-2\frac{dy}{dx}\times \left(2y+2\right)^{-2}
\frac{d^2y}{dx^2}=-2\frac{dy}{dx}\times \left(2y+2\right)^{-2}
Rewrite the expression
\frac{d^2y}{dx^2}=-\frac{2\frac{dy}{dx}}{\left(2y+2\right)^{2}}
Calculate
\frac{d^2y}{dx^2}=-\frac{\frac{dy}{dx}}{2\left(y+1\right)^{2}}
\text{Use equation }\frac{dy}{dx}=\frac{1}{2y+2}\text{ to substitute}
\frac{d^2y}{dx^2}=-\frac{\frac{1}{2y+2}}{2\left(y+1\right)^{2}}
Solution
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Calculate
-\frac{\frac{1}{2y+2}}{2\left(y+1\right)^{2}}
Divide the terms
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Evaluate
\frac{\frac{1}{2y+2}}{2\left(y+1\right)^{2}}
Multiply by the reciprocal
\frac{1}{2y+2}\times \frac{1}{2\left(y+1\right)^{2}}
Multiply the terms
\frac{1}{\left(2y+2\right)\times 2\left(y+1\right)^{2}}
Use the commutative property to reorder the terms
\frac{1}{2\left(2y+2\right)\left(y+1\right)^{2}}
-\frac{1}{2\left(2y+2\right)\left(y+1\right)^{2}}
Expand the expression
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Evaluate
2\left(2y+2\right)\left(y+1\right)^{2}
Expand the expression
2\left(2y+2\right)\left(y^{2}+2y+1\right)
Multiply the terms
\left(4y+4\right)\left(y^{2}+2y+1\right)
Apply the distributive property
4y\times y^{2}+4y\times 2y+4y\times 1+4y^{2}+4\times 2y+4\times 1
Multiply the terms
4y^{3}+4y\times 2y+4y\times 1+4y^{2}+4\times 2y+4\times 1
Multiply the terms
4y^{3}+8y^{2}+4y\times 1+4y^{2}+4\times 2y+4\times 1
Any expression multiplied by 1 remains the same
4y^{3}+8y^{2}+4y+4y^{2}+4\times 2y+4\times 1
Multiply the numbers
4y^{3}+8y^{2}+4y+4y^{2}+8y+4\times 1
Any expression multiplied by 1 remains the same
4y^{3}+8y^{2}+4y+4y^{2}+8y+4
Add the terms
4y^{3}+12y^{2}+4y+8y+4
Add the terms
4y^{3}+12y^{2}+12y+4
-\frac{1}{4y^{3}+12y^{2}+12y+4}
\frac{d^2y}{dx^2}=-\frac{1}{4y^{3}+12y^{2}+12y+4}
Show Solutions
Rewrite the equation
\begin{align}&r=\frac{\cos\left(\theta \right)-2\sin\left(\theta \right)-\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{2\sin^{2}\left(\theta \right)}\\&r=\frac{\cos\left(\theta \right)-2\sin\left(\theta \right)+\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{2\sin^{2}\left(\theta \right)}\end{align}
Evaluate
x=y^{2}+2y+2
Move the expression to the left side
x-y^{2}-2y=2
\text{To convert the equation to polar coordinates,substitute }r\cos\left(\theta \right)\text{ for }x\text{ and }r\sin\left(\theta \right)\text{ for }y
\cos\left(\theta \right)\times r-\left(\sin\left(\theta \right)\times r\right)^{2}-2\sin\left(\theta \right)\times r=2
Factor the expression
-\sin^{2}\left(\theta \right)\times r^{2}+\left(\cos\left(\theta \right)-2\sin\left(\theta \right)\right)r=2
Subtract the terms
-\sin^{2}\left(\theta \right)\times r^{2}+\left(\cos\left(\theta \right)-2\sin\left(\theta \right)\right)r-2=2-2
Evaluate
-\sin^{2}\left(\theta \right)\times r^{2}+\left(\cos\left(\theta \right)-2\sin\left(\theta \right)\right)r-2=0
Solve using the quadratic formula
r=\frac{-\cos\left(\theta \right)+2\sin\left(\theta \right)\pm \sqrt{\left(\cos\left(\theta \right)-2\sin\left(\theta \right)\right)^{2}-4\left(-\sin^{2}\left(\theta \right)\right)\left(-2\right)}}{-2\sin^{2}\left(\theta \right)}
Simplify
r=\frac{-\cos\left(\theta \right)+2\sin\left(\theta \right)\pm \sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{-2\sin^{2}\left(\theta \right)}
\text{Separate the equation into }2\text{ possible cases}
\begin{align}&r=\frac{-\cos\left(\theta \right)+2\sin\left(\theta \right)+\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{-2\sin^{2}\left(\theta \right)}\\&r=\frac{-\cos\left(\theta \right)+2\sin\left(\theta \right)-\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{-2\sin^{2}\left(\theta \right)}\end{align}
\text{Use }\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}\text{ to rewrite the fraction}
\begin{align}&r=\frac{\cos\left(\theta \right)-2\sin\left(\theta \right)-\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{2\sin^{2}\left(\theta \right)}\\&r=\frac{-\cos\left(\theta \right)+2\sin\left(\theta \right)-\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{-2\sin^{2}\left(\theta \right)}\end{align}
Solution
\begin{align}&r=\frac{\cos\left(\theta \right)-2\sin\left(\theta \right)-\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{2\sin^{2}\left(\theta \right)}\\&r=\frac{\cos\left(\theta \right)-2\sin\left(\theta \right)+\sqrt{5\cos^{2}\left(\theta \right)-2\sin\left(2\theta \right)-4}}{2\sin^{2}\left(\theta \right)}\end{align}
Show Solutions
Graph
