Algebra
Calculus
Trigonometry
Matrix
Question
Function
Find the vertex
Find the axis of symmetry
Rewrite in vertex form
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(211,−4241)
Evaluate
y=x2−11x−30
Find the x-coordinate of the vertex by substituting a=1 and b=−11 into x = −2ab
x=−2×1−11
Solve the equation for x
x=211
Find the y-coordinate of the vertex by evaluating the function for x=211
y=(211)2−11×211−30
Calculate
More Steps
Evaluate
(211)2−11×211−30
Multiply the numbers
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Evaluate
−11×211
Multiply the numbers
−211×11
Multiply the numbers
−2121
(211)2−2121−30
Evaluate the power
4121−2121−30
Reduce fractions to a common denominator
4121−2×2121×2−2×230×2×2
Multiply the numbers
4121−4121×2−2×230×2×2
Multiply the numbers
4121−4121×2−430×2×2
Write all numerators above the common denominator
4121−121×2−30×2×2
Multiply the numbers
4121−242−30×2×2
Multiply the terms
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Evaluate
30×2×2
Multiply the terms
60×2
Multiply the numbers
120
4121−242−120
Subtract the numbers
4−241
Use b−a=−ba=−ba to rewrite the fraction
−4241
y=−4241
Solution
(211,−4241)
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=x2−11x−30
To test if the graph of y=x2−11x−30 is symmetry with respect to the origin,substitute -x for x and -y for y
−y=(−x)2−11(−x)−30
Simplify
More Steps
Evaluate
(−x)2−11(−x)−30
Multiply the numbers
(−x)2+11x−30
Rewrite the expression
x2+11x−30
−y=x2+11x−30
Change the signs both sides
y=−x2−11x+30
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−211)2=y+4241
Evaluate
y=x2−11x−30
Swap the sides of the equation
x2−11x−30=y
Move the constant to the right-hand side and change its sign
x2−11x=y−(−30)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
x2−11x=y+30
To complete the square, the same value needs to be added to both sides
x2−11x+4121=y+30+4121
Use a2−2ab+b2=(a−b)2 to factor the expression
(x−211)2=y+30+4121
Solution
More Steps
Evaluate
30+4121
Reduce fractions to a common denominator
430×4+4121
Write all numerators above the common denominator
430×4+121
Multiply the numbers
4120+121
Add the numbers
4241
(x−211)2=y+4241
Show Solution
Solve the equation
x=211+241+4yx=211−241+4y
Evaluate
y=x2−11x−30
Swap the sides of the equation
x2−11x−30=y
Move the expression to the left side
x2−11x−30−y=0
Substitute a=1,b=−11 and c=−30−y into the quadratic formula x=2a−b±b2−4ac
x=211±(−11)2−4(−30−y)
Simplify the expression
More Steps
Evaluate
(−11)2−4(−30−y)
Multiply the terms
More Steps
Evaluate
4(−30−y)
Apply the distributive property
−4×30−4y
Multiply the numbers
−120−4y
(−11)2−(−120−4y)
Rewrite the expression
112−(−120−4y)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
112+120+4y
Evaluate the power
121+120+4y
Add the numbers
241+4y
x=211±241+4y
Solution
x=211+241+4yx=211−241+4y
Show Solution
Rewrite the equation
r=2cos2(θ)sin(θ)+11cos(θ)−1+240cos2(θ)+11sin(2θ)r=2cos2(θ)sin(θ)+11cos(θ)+1+240cos2(θ)+11sin(2θ)
Evaluate
y=x2−11x−30
Move the expression to the left side
y−x2+11x=−30
To convert the equation to polar coordinates,substitute x for rcos(θ) and y for rsin(θ)
sin(θ)×r−(cos(θ)×r)2+11cos(θ)×r=−30
Factor the expression
−cos2(θ)×r2+(sin(θ)+11cos(θ))r=−30
Subtract the terms
−cos2(θ)×r2+(sin(θ)+11cos(θ))r−(−30)=−30−(−30)
Evaluate
−cos2(θ)×r2+(sin(θ)+11cos(θ))r+30=0
Solve using the quadratic formula
r=−2cos2(θ)−sin(θ)−11cos(θ)±(sin(θ)+11cos(θ))2−4(−cos2(θ))×30
Simplify
r=−2cos2(θ)−sin(θ)−11cos(θ)±1+240cos2(θ)+11sin(2θ)
Separate the equation into 2 possible cases
r=−2cos2(θ)−sin(θ)−11cos(θ)+1+240cos2(θ)+11sin(2θ)r=−2cos2(θ)−sin(θ)−11cos(θ)−1+240cos2(θ)+11sin(2θ)
Use b−a=−ba=−ba to rewrite the fraction
r=2cos2(θ)sin(θ)+11cos(θ)−1+240cos2(θ)+11sin(2θ)r=−2cos2(θ)−sin(θ)−11cos(θ)−1+240cos2(θ)+11sin(2θ)
Solution
r=2cos2(θ)sin(θ)+11cos(θ)−1+240cos2(θ)+11sin(2θ)r=2cos2(θ)sin(θ)+11cos(θ)+1+240cos2(θ)+11sin(2θ)
Show Solution
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