Create a system of equations and use algebra to write a quadratic function that passes through the points (–2, 8), (1, 14), and (0, 10).

y = x² + 3x + 10

the equation of a parabola in standard form is → y = ax² + bx + c ( a ≠ 0 )

Substitute each of the pairs of coordinate points into the equation and solve for a, b and c

(- 2, 8) → 8 = 4a + 2b + c → (1)

(1, 14) → 14 = a + b + c → (2)

note that (0 , 10) is the y-intercept ⇒ c = 10

substitute c = 10 into (1) and (2)

8 = 4a - 2b + 10 → 4a - 2b = - 2 → (3)

14 = a + b + 10 → a + b = 4 → (4)

multiply (4 ) by 2

2a + 2b = 8 → (5)

add (3) and (5) term by term ⇒ 6a = 6 ⇒ a = 1

substitute a = 1 into (4)

1 + b = 4 ⇒ b = 3

thus a = 1, b= 3 and c = 10

equation of parabola is y = x² + 3x + 10

it is probably the graph on the bottom right.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1