Quadratic: finding the rule, word problems |
Quadratic Function (second degree)
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Quadratic Function Rule Templates
standard form | general form | zeros form |
f (x) = a( x – h )^{ 2} + k given (h, k), point (x_{1}, y_{1}) plug in (x_{1}, y_{1}) to find a. |
f (x)= ax^{2} + bx + c c is the y-intercept factor trinomial for zeros use h = – b/2a and k = f (h) to change to standard form. |
f (x) = a(x – x_{1})( x – x_{2}) x_{1} and x_{2} are zeros. x = h is ½ way between. k = f (h) |
Example 1
Find the rule of correspondence in all 3 forms for a parabola with vertex at (1, –9) and
y-intercept of – 8.
Solution:
Since we know the vertex, we'll start with the standard or function form.
f (x) = a( x –h )^{ 2} + k f (x) = a( x –1 )^{ 2} –9
Now we plug in the coordinates of the y-intercept (0, –8) to get a
–8 = a( 0 –1 )^{ 2} –9 a = 1 So f (x) = ( x – 1 )² – 9
To get the general form, expand the brackets and collect terms.
f (x) = ( x –1 )^{ 2} –9 f (x) = x^{ }2 – 2x – 8
To get the zeros form, factor the trinomial.
f (x) = (x –4)(x + 2) so the zeros are x = 4 and x = –2
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Example 2
The cross section of a riverbed is a parabola, 60 meters wide with a maximum depth of
6 meters. At what distance from the shores must we place 2 buoys to mark where
the depth is 4 meters? (Hint: use the y-axis as the axis of symmetry.)
Solution:
Since the river is 60 m wide, each shore is 30 m from the vertex or deepest point.
This puts the zeros at (–30, 0) and (30, 0)
Since the lowest point (maximum depth) is 6 m below the surface,
the vertex is at (0, – 6).
Now we find the rule of correspondence.
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Hint: With any type of function, once we have the rule of correspondence, the only thing left to figure out is x-values when we're given y-values, or vice versa.
In the previous example that's what we had. We wanted the x's that go with y = – 4.
Sometimes, we get x-values and we have to find the corresponding y's.
reminder: f (x) is just an alias for y, so we substitute the given values for x in the rule and calculate the resulting y-value.
.Practice
1) Find the rule of correspondence for a parabola:
a) with vertex at (1, 1) and y-intercept of 5.
b) with zeros at – 1 and 3, passing through P(7, – 16). (click for solution)
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2) Find the zeros for these parabolas.
a) f (x) = 2x^{ 2} + 5x – 3
b) f (x) = – (x + 1)^{ 2} + 4
c) f (x) = – 5x^{ 2} + 4x + 1
d) f (x) = 2(x^{ 2} –8x + 17) (click for solution)
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3) Word problems with quadratics: (make a diagram where needed!)
a) A room measures 5 m by 6 m. A rectangular rug leaves an exposed border of equal width on all sides, and covers of the floor area. What are the dimensions of the rug? (click for solution)
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b) El Cheapo High School awarded a total of $1440 in prize money to students
at the end of the year. If there had been 10 less winners, each prize would be $2 greater.
How many students received prize money? (click for solution)
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c) The initial price of a share of Miscumbooberated Products was $8. It fell for 25 months and reached a minimum of $3. Then it began to rise. If the price of a share follows a parabolic curve, how many months will it take for it to be $6 for the second time? (diagram!!)
(click for solution)
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d) Raymond rents cabins during ice fishing season.
In 1997 he charged $75/day and had an average of 50 customers per day.
In 1998, Raymond decided to lower the daily rental rate by x dollars and
he expects that the number of daily customers will rise by twice the value of x.
What is the maximum daily income that Raymond can earn in 1998 with the new conditions?
(click for solution).
e) The cross-section of a sculpture is depicted in the diagram.
The equation of the parabola is y = – x^{ }² + 36x – 284.
The equation of the line BC is y = 3x – 24.
What is the measure of segment AB? (click for solution)
Solutions
1) a) vertex (1, 1), y-intercept 5.
f (x) = a( x – h )^{ }² + k f (x) = a( x – 1 )² + 1
Now we plug in the coordinates of the y-intercept (0, 5) to get 5 = a( 0 – 1 )² + 1 a = 4
So f (x) = 4( x – 1 )² + 1
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b) with zeros at – 1 and 3, passing through P(7, – 16).
f (x) = a(x – x_{1} )( x – x_{2}) f (x) = a(x + 1)( x – 3)
Now we plug in the coordinates of P(7, – 16)
–16 = a(7 + 1)( 7 –3) a = –½
f (x) = –½ (x+ 1)( x – 3)
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2) a) f (x) = 2x²^{ }+ 5x – 3
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b) f (x) = – (x + 1)^{ 2} + 4
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c) f (x) = – 5x^{ 2} + 4x + 1
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d) f (x) = 2(x^{ 2} – 8x + 17)
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b) Let x = # of students who won a prize
Multiply through by the lcd x(x – 10) to get
1440(x – 10) + 2x(x – 10) = 1440x
x^{ 2} – 10x – 7200 = 0 (x – 90)( x + 80) = 0
Ninety (90) students shared the prize money.
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c) We know that (0, 8) is the y-intercept and (25, 3) is the vertex.
Set up the standard form template:
f (x) = a( x – h )^{ 2} + k f (x) = a( x – 25 )^{ 2} + 3
Now substitute (0, 8) for x and y to get: 8 = a( 0 –25 )^{ 2} + 3
So a = 1/125 which makes f (x) = 1/125 ( x – 25 )^{ 2} + 3
We need the larger of 2 x-values that make f (x) or y = 6 so set y = 6.
6 = 1/125 ( x – 25 )^{ 2} + 3 375 = ( x – 25 )^{ 2} so x = 44.36 or 45 months
1977 | 1978 | ||
rent ($) | # of renters | rent ($) | # of renters |
75 | 50/day | (75 – x) | (50 + 2x) |
The income function is I(x) = (75 – x)(50 + 2x). We want h, the x-value of the vertex.
We can see that the zeros are 75 and – 25, and we know h is half way between
so h = 25 = x. This means the max income happens when x = 25 and f (x) = $5000
Raymond's maximum daily income will be $5000 per day if he lowers the rent by $25.
e) What we need here is the y-value of point B, the 2nd point of intersection.
Let's find the points of intersection for the parabola and the line:
y = – x ² + 36x – 284 and y = 3x – 24 3x – 24= – x² + 36x – 284 or x² – 33x + 260 = 0
This trinomial factors as: (x – 20)(x – 13) = 0. Since we want the larger one, x = 20.
Substituting into y = 3x – 24, we get y = 36. AB = 36.
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