Verified Answer :

djsdoodle

Plug in points to general form of quadratic function:
y=ax²+bx+c

(0,7)
7 = c
(2,13)
13 = 4a +2b + c
(4,3)
3 = 16a + 4b + c

Now solve the system of equations in two unknowns.

Substitute the value of 7 for c in the second equation:
13 = 4a + 2b + 7
6 = 4a + 2b
3 = 2a + b
b = 3 - 2a

Substitute this into third equation:
3 = 16a + 4(3-2a) + 7
3 = 16a + 12 - 8a + 7
3 = 8a + 19
-16 = 8a
a = -2

b = 3-2a = 3 - 2(-2) = 7

a = -2, b =7. c = 7
The quadratic function is y= -2x² + 7x +7

yoodyannapolis

The quadratic function is \(y= -2x^2 + 7x +7\)

The standard quadratic function is

\(y=ax^2+bx+c\)


We have set of values (0,7),(2, 13), (4, 3)

Substitute these values in the standard quadratic function and find the unknown a, b and c.

Put (0,7) in quadratic function, we get

7 =a.0+b.0+ c

c=7

Put (2,13) in quadratic function, we get

13 = 4a +2b + c...(2)

Put (4,3) in quadratic function, we get

3 = 16a + 4b + c...(3)


Now, solving the system of equations


Substitute c=7 in the second equation, we get

13 = 4a + 2b + 7

6 = 4a + 2b

3 = 2a + b

b = 3 - 2a


Substitute b = 3 - 2a in (3), we get

\(3 = 16a + 4(3-2a) + 7\\3 = 16a + 12 - 8a + 7\\3 = 8a + 19\\-16 = 8a\\a = -2\\b = 3-2a \\= 3 - 2(-2)\\ = 7\\a = -2, b =7. c = 7\)

Therefore the quadratic function is \(y= -2x^2 + 7x +7\)

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