Question
Function
Find the vertex
Find the axis of symmetry
Rewrite in vertex form
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(25,−481)
Evaluate
y=x2−5x−14
Find the x-coordinate of the vertex by substituting a=1 and b=−5 into x = −2ab
x=−2×1−5
Solve the equation for x
x=25
Find the y-coordinate of the vertex by evaluating the function for x=25
y=(25)2−5×25−14
Calculate
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Evaluate
(25)2−5×25−14
Multiply the numbers
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Evaluate
−5×25
Multiply the numbers
−25×5
Multiply the numbers
−225
(25)2−225−14
Evaluate the power
425−225−14
Reduce fractions to a common denominator
425−2×225×2−2×214×2×2
Multiply the numbers
425−425×2−2×214×2×2
Multiply the numbers
425−425×2−414×2×2
Write all numerators above the common denominator
425−25×2−14×2×2
Multiply the numbers
425−50−14×2×2
Multiply the terms
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Evaluate
14×2×2
Multiply the terms
28×2
Multiply the numbers
56
425−50−56
Subtract the numbers
4−81
Use b−a=−ba=−ba to rewrite the fraction
−481
y=−481
Solution
(25,−481)
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=x2−5x−14
To test if the graph of y=x2−5x−14 is symmetry with respect to the origin,substitute -x for x and -y for y
−y=(−x)2−5(−x)−14
Simplify
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Evaluate
(−x)2−5(−x)−14
Multiply the numbers
(−x)2+5x−14
Rewrite the expression
x2+5x−14
−y=x2+5x−14
Change the signs both sides
y=−x2−5x+14
Solution
Not symmetry with respect to the origin
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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−25)2=y+481
Evaluate
y=x2−5x−14
Swap the sides of the equation
x2−5x−14=y
Move the constant to the right-hand side and change its sign
x2−5x=y−(−14)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
x2−5x=y+14
To complete the square, the same value needs to be added to both sides
x2−5x+425=y+14+425
Use a2−2ab+b2=(a−b)2 to factor the expression
(x−25)2=y+14+425
Solution
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Evaluate
14+425
Reduce fractions to a common denominator
414×4+425
Write all numerators above the common denominator
414×4+25
Multiply the numbers
456+25
Add the numbers
481
(x−25)2=y+481
Show Solution
Solve the equation
Solve for x
x=281+4y+5x=2−81+4y+5
Evaluate
y=x2−5x−14
Swap the sides of the equation
x2−5x−14=y
Move the expression to the left side
x2−5x−14−y=0
Move the constant to the right side
x2−5x=0−(−14−y)
Add the terms
x2−5x=14+y
Add the same value to both sides
x2−5x+425=14+y+425
Evaluate
x2−5x+425=481+4y
Evaluate
(x−25)2=481+4y
Take the root of both sides of the equation and remember to use both positive and negative roots
x−25=±481+4y
Simplify the expression
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Evaluate
481+4y
To take a root of a fraction,take the root of the numerator and denominator separately
481+4y
Simplify the radical expression
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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
281+4y
x−25=±281+4y
Separate the equation into 2 possible cases
x−25=281+4yx−25=−281+4y
Calculate
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Evaluate
x−25=281+4y
Move the constant to the right-hand side and change its sign
x=281+4y+25
Write all numerators above the common denominator
x=281+4y+5
x=281+4y+5x−25=−281+4y
Solution
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Evaluate
x−25=−281+4y
Move the constant to the right-hand side and change its sign
x=−281+4y+25
Write all numerators above the common denominator
x=2−81+4y+5
x=281+4y+5x=2−81+4y+5
Show Solution
Rewrite the equation
Rewrite in polar form
r=2cos2(θ)sin(θ)+5cos(θ)−1+80cos2(θ)+5sin(2θ)r=2cos2(θ)sin(θ)+5cos(θ)+1+80cos2(θ)+5sin(2θ)
Evaluate
y=x2−5x−14
Move the expression to the left side
y−x2+5x=−14
To convert the equation to polar coordinates,substitute rcos(θ) for x and rsin(θ) for y
sin(θ)×r−(cos(θ)×r)2+5cos(θ)×r=−14
Factor the expression
−cos2(θ)×r2+(sin(θ)+5cos(θ))r=−14
Subtract the terms
−cos2(θ)×r2+(sin(θ)+5cos(θ))r−(−14)=−14−(−14)
Evaluate
−cos2(θ)×r2+(sin(θ)+5cos(θ))r+14=0
Solve using the quadratic formula
r=−2cos2(θ)−sin(θ)−5cos(θ)±(sin(θ)+5cos(θ))2−4(−cos2(θ))×14
Simplify
r=−2cos2(θ)−sin(θ)−5cos(θ)±1+80cos2(θ)+5sin(2θ)
Separate the equation into 2 possible cases
r=−2cos2(θ)−sin(θ)−5cos(θ)+1+80cos2(θ)+5sin(2θ)r=−2cos2(θ)−sin(θ)−5cos(θ)−1+80cos2(θ)+5sin(2θ)
Use b−a=−ba=−ba to rewrite the fraction
r=2cos2(θ)sin(θ)+5cos(θ)−1+80cos2(θ)+5sin(2θ)r=−2cos2(θ)−sin(θ)−5cos(θ)−1+80cos2(θ)+5sin(2θ)
Solution
r=2cos2(θ)sin(θ)+5cos(θ)−1+80cos2(θ)+5sin(2θ)r=2cos2(θ)sin(θ)+5cos(θ)+1+80cos2(θ)+5sin(2θ)
Show Solution