Algebra
Calculus
Trigonometry
Matrix
Question
Function
Find the vertex
Find the axis of symmetry
Rewrite in vertex form
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(−31,32)
Evaluate
y=3x2+2x+1
Find the x-coordinate of the vertex by substituting a=3 and b=2 into x = −2ab
x=−2×32
Solve the equation for x
x=−31
Find the y-coordinate of the vertex by evaluating the function for x=−31
y=3(−31)2+2(−31)+1
Calculate
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Evaluate
3(−31)2+2(−31)+1
Multiply the terms
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Evaluate
3(−31)2
Evaluate the power
3×91
Multiply the numbers
31
31+2(−31)+1
Multiply the numbers
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Evaluate
2(−31)
Multiplying or dividing an odd number of negative terms equals a negative
−2×31
Multiply the numbers
−32
31−32+1
Reduce fractions to a common denominator
31−32+33
Write all numerators above the common denominator
31−2+3
Calculate the sum or difference
32
y=32
Solution
(−31,32)
Show Solution
Testing for symmetry
Testing for symmetry about the origin
Testing for symmetry about the x-axis
Testing for symmetry about the y-axis
Not symmetry with respect to the origin
Evaluate
y=3x2+2x+1
To test if the graph of y=3x2+2x+1 is symmetry with respect to the origin,substitute -x for x and -y for y
−y=3(−x)2+2(−x)+1
Simplify
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Evaluate
3(−x)2+2(−x)+1
Multiply the terms
3x2+2(−x)+1
Multiply the numbers
3x2−2x+1
−y=3x2−2x+1
Change the signs both sides
y=−3x2+2x−1
Solution
Not symmetry with respect to the origin
Show Solution
Identify the conic
Find the standard equation of the parabola
Find the vertex of the parabola
Find the focus of the parabola
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(x+31)2=31(y−32)
Evaluate
y=3x2+2x+1
Swap the sides of the equation
3x2+2x+1=y
Move the constant to the right-hand side and change its sign
3x2+2x=y−1
Multiply both sides of the equation by 31
(3x2+2x)×31=(y−1)×31
Multiply the terms
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Evaluate
(3x2+2x)×31
Use the the distributive property to expand the expression
3x2×31+2x×31
Multiply the numbers
x2+2x×31
Multiply the numbers
x2+32x
x2+32x=(y−1)×31
Multiply the terms
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Evaluate
(y−1)×31
Apply the distributive property
y×31−31
Use the commutative property to reorder the terms
31y−31
x2+32x=31y−31
To complete the square, the same value needs to be added to both sides
x2+32x+91=31y−31+91
Use a2+2ab+b2=(a+b)2 to factor the expression
(x+31)2=31y−31+91
Add the numbers
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Evaluate
−31+91
Reduce fractions to a common denominator
−3×33+91
Multiply the numbers
−93+91
Write all numerators above the common denominator
9−3+1
Add the numbers
9−2
Use b−a=−ba=−ba to rewrite the fraction
−92
(x+31)2=31y−92
Solution
(x+31)2=31(y−32)
Show Solution
Solve the equation
x=3−1+−2+3yx=−31+−2+3y
Evaluate
y=3x2+2x+1
Swap the sides of the equation
3x2+2x+1=y
Move the expression to the left side
3x2+2x+1−y=0
Substitute a=3,b=2 and c=1−y into the quadratic formula x=2a−b±b2−4ac
x=2×3−2±22−4×3(1−y)
Simplify the expression
x=6−2±22−4×3(1−y)
Simplify the expression
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Evaluate
22−4×3(1−y)
Multiply the terms
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Multiply the terms
4×3(1−y)
Multiply the terms
12(1−y)
Apply the distributive property
12−12y
22−(12−12y)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
22−12+12y
Evaluate the power
4−12+12y
Subtract the numbers
−8+12y
x=6−2±−8+12y
Simplify the radical expression
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Evaluate
−8+12y
Factor the expression
4(−2+3y)
The root of a product is equal to the product of the roots of each factor
4×−2+3y
Evaluate the root
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Evaluate
4
Write the number in exponential form with the base of 2
22
Reduce the index of the radical and exponent with 2
2
2−2+3y
x=6−2±2−2+3y
Separate the equation into 2 possible cases
x=6−2+2−2+3yx=6−2−2−2+3y
Simplify the expression
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Evaluate
x=6−2+2−2+3y
Divide the terms
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Evaluate
6−2+2−2+3y
Rewrite the expression
62(−1+−2+3y)
Cancel out the common factor 2
3−1+−2+3y
x=3−1+−2+3y
x=3−1+−2+3yx=6−2−2−2+3y
Solution
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Evaluate
x=6−2−2−2+3y
Divide the terms
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Evaluate
6−2−2−2+3y
Rewrite the expression
62(−1−−2+3y)
Cancel out the common factor 2
3−1−−2+3y
Use b−a=−ba=−ba to rewrite the fraction
−31+−2+3y
x=−31+−2+3y
x=3−1+−2+3yx=−31+−2+3y
Show Solution
Rewrite the equation
r=6cos2(θ)sin(θ)−2cos(θ)−1−9cos2(θ)−2sin(2θ)r=6cos2(θ)sin(θ)−2cos(θ)+1−9cos2(θ)−2sin(2θ)
Evaluate
y=3x2+2x+1
Move the expression to the left side
y−3x2−2x=1
To convert the equation to polar coordinates,substitute x for rcos(θ) and y for rsin(θ)
sin(θ)×r−3(cos(θ)×r)2−2cos(θ)×r=1
Factor the expression
−3cos2(θ)×r2+(sin(θ)−2cos(θ))r=1
Subtract the terms
−3cos2(θ)×r2+(sin(θ)−2cos(θ))r−1=1−1
Evaluate
−3cos2(θ)×r2+(sin(θ)−2cos(θ))r−1=0
Solve using the quadratic formula
r=−6cos2(θ)−sin(θ)+2cos(θ)±(sin(θ)−2cos(θ))2−4(−3cos2(θ))(−1)
Simplify
r=−6cos2(θ)−sin(θ)+2cos(θ)±1−9cos2(θ)−2sin(2θ)
Separate the equation into 2 possible cases
r=−6cos2(θ)−sin(θ)+2cos(θ)+1−9cos2(θ)−2sin(2θ)r=−6cos2(θ)−sin(θ)+2cos(θ)−1−9cos2(θ)−2sin(2θ)
Use b−a=−ba=−ba to rewrite the fraction
r=6cos2(θ)sin(θ)−2cos(θ)−1−9cos2(θ)−2sin(2θ)r=−6cos2(θ)−sin(θ)+2cos(θ)−1−9cos2(θ)−2sin(2θ)
Solution
r=6cos2(θ)sin(θ)−2cos(θ)−1−9cos2(θ)−2sin(2θ)r=6cos2(θ)sin(θ)−2cos(θ)+1−9cos2(θ)−2sin(2θ)
Show Solution
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