Question

Find the values of​ a, b, and c such that the equation y=ax2+bx+c has ordered pair solutions ​(1​,16​), ​(−1​,−2​), and ​(0​,−1​). To do​ so, substitute each ordered pair solution into the equation. Each​ time, the result is an equation in three​ unknowns, a,​ b, and c. Then solve the resulting system of three linear equations in three​ unknowns, a,​ b, and c.

Answer 1

c = -1

b = 9

a = 8

Explanation:

16 = a × 1² + b × 1 + c

-2 = a × (-1)² + b × (-1) + c

-1 = a × 0² + b × 0 + c .........-> -1 = c

16 = a + b + c

-2 = a - b + c

16 = a + b - 1

-2 = a - b - 1

a + b = 17

a - b = -1 .......-> a = b - 1 ........ a = 9 - 1 = 8

(b - 1) + b = 17

2b - 1 = 17

2b = 18 .......-> b = 18/2 = 9

Answer 2

y = 8x² +9x -1

Explanation:

The coefficients of the quadratic function can be found from three points by substituting the point values into the equation and solving for the coefficients.

The equation we want to find coefficients for is ...

y = ax² +bx +c

For point (1, 16), the equation is 16 = a(1²) +b(1) +c

For point (-1, -2), the equation is -2 = a(-1)² +b(-1) +c

For point (0, -1), the equation is -1 = a(0) +b(0) +c

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The last of these equations gives a value for c (c=-1). Substituting that into the other two gives the equations in a and b as ...

a + b = 17

a - b = -1

Adding the two equations gives ...

2a = 16 ⇒ a = 8

Subtracting the second from the first gives ...

2b = 18 ⇒ b = 9

Then the desired quadratic equation is ...

y = 8x² +9x -1