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Quadratic function passing through two points

by Wayne
(Abbotsford, Canada)










































Equation of Quadratic Function

Determine the equation of the quadratic function with the given characteristics of it's graph:

x-y coordinates of the vertex: (5,4)

y-intercept 79

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Oct 23, 2013
Equation of Quadratic Function
by: Staff


Answer

Part I

The standard form of a quadratic function (a parabola) is:

y = ax² + bx + c



Quadratic Function - standard form (parabola)






(i) the value of the constant “c”


The quadratic function (a parabola) passes through the y-intercept (0,79).

When x = 0, y = 79


To solve for the coefficient “c”, substitute 0 for x and 79 for y in the standard form of the quadratic equation.

y = ax² + bx + c

79 = a*0² + b*0 + c

79 = 0 + 0 + c

79 = c

c = 79




Quadratic Function - use the y-intercept to solve for the 'c' coefficient




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Oct 23, 2013
Equation of Quadratic Function
by: Staff


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Part II



Quadratic Function - substitute 0 for x and 79 for y, and then solve for the 'c' coefficient





(ii) the value of the coefficient “a”



we know the x and y coordinates at the vertex



x-y coordinates of the vertex: (5,4)





Parabola - coordinates of the vertex








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Oct 23, 2013
Equation of Quadratic Function
by: Staff


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Part III




substituting these values in the quadratic function, we have an equation with two unknowns (the coefficients a and b)



y = ax² + bx + c

x = 5

y = 4

c = 79



4 = a*5² + b*5 + 79

4 = 25a + 5b + 79




substitute the following values in the equation for the parabola, and then simplify the result

        x = 5

        y = 4

        c = 79





However, we also know that the first term in the quadratic formula is the x coordinate of the vertex.

Not only that, but it contains the same two coefficients: "a" and "b"



x coordinate of the vertex = -b/2a

x = -b/2a




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Oct 23, 2013
Equation of Quadratic Function
by: Staff


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Part IV



the first term of the quadratic formula is equivalent to the x coordinate of the vertex of the parabola






substituting the known value for x (x = 5 at the vertex)



5 = -b/2a


solve for b

multiply each side of the equation by (-1)



(-1)*5 = (-1)*(-b/2a)


-5 = b/2a


multiply each side of the equation by 2a



(2a)*(-5) = (2a)*(b/2a)

-10a = b

b = -10a




substitute 5 for x in the equation







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Oct 23, 2013
Equation of Quadratic Function
by: Staff


-------------------------------------------------




Part V



solve the equation for the 'b' coefficient in terms of 'a'






substitute -10a for b in the equation with the two unknowns



4 = 25a + 5b + 79

4 = 25a + 5*(-10a) + 79

4 = 25a - 50a + 79

4 = -25a + 79





substitute '-10a' for the coefficient 'b' in the first  equation with the two unknowns






add 25a to each side of the equation and combine like terms



4 + 25a = -25a + 79 + 25a

4 + 25a = -25a + 25a + 79

4 + 25a = 0 + 79

4 + 25a = 79






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Oct 23, 2013
Equation of Quadratic Function
by: Staff


-------------------------------------------------




Part VI



solve for the coefficient 'a'






subtract 4 from each side of the equation



4 + 25a - 4 = 79 - 4

4 - 4 + 25a= 79 - 4

0 + 25a = 79 - 4

25a = 79 - 4

25a = 75





subtract 4 from each side of the equation






divide each side of the equation by 25



25a/25 = 75/25

(25/25)a = 75/25

(1)a = 75/25

a = 75/25

a = 3







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Oct 23, 2013
Equation of Quadratic Function
by: Staff


-------------------------------------------------




Part VII




divide each side of the equation by 25






(iii) the value of the coefficient “b”



from section (ii) we determined that



b = -10a


since we computed the value of the coefficient "a" in section (ii), substitute that value for "a"



b = -10a

a = 3

b = -10 * 3

b = -30




substitute 3 for the coefficient 'a' in the equation








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Oct 23, 2013
Equation of Quadratic Function
by: Staff


-------------------------------------------------




Part VIII



(iv) the final equation



y = ax² + bx + c



a = 3

b = -30

c = 79


y = 3*x² + (-30)*x + 79



y = 3x² - 30x + 79




the final equation, showing the following coefficients: 

        a = 3

        b = -30

        c = 79






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(v) check your work



x-y coordinates of the vertex: (5,4)



if the equation is correct, when x = 5, then y should = 4


substitute 5 for x and 4 for y in the final equation



y = 3x² - 30x + 79

y = 3*5² - 30*5 + 79

y = 3*25 - 150 + 79

y = 75 - 150 + 79

y = 4, OK



y-intercept = 79




if the equation is correct, when x = 0, then y should = 79


substitute 0 for x and 79 for y in the final equation



y = 3x² - 30x + 79

y = 3*0² - 30*0 + 79

y = 0 - 0 + 79

y = 79, OK





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Oct 23, 2013
Equation of Quadratic Function
by: Staff


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Part IX



A graph of the equation of the quadratic function is shown below



graph of the final quadratic function: 

        y = 3x² - 30x + 79








Thanks for writing.

Staff
www.solving-math-problems.com


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