Question

4. Find “a” so that the parabola y = ax² + 3x + 11 goes through the point (3,2).

Answer 1

a = - 2

Explanation:

since the point (3, 2 ) lies on the parabola, then it makes the equation true.

substitute x = 3, y = 2 into the equation and solve for a

y = ax² + 3x + 11

2 = a(3)² + 3(3) + 11

2 = 9a + 9 + 11

2 = 9a + 20 ( subtract 20 from both sides )

- 18 = 9a ( divide both sides by 9 )

- 2 = a

Answer 2

a = -2

Explanation:

Given quadratic equation:

`y=ax^2+3x+11`

In order to find "a", substitute the given point on the curve (3, 2) into the equation and solve for a:

`\begin{aligned}y&=ax^2+3x+11\\\textsf{Substitute $(3,2)$} \implies 2&=a(3)^2+3(3)+11\\2&=9a+9+11\\2&=9a+20\\2-20&=9a+20-20\\-18&=9a\\9a&=-18\\(9a)/(9)&=(-18)/(9)\\\implies a&=-2 \end{aligned}`

Therefore, the value of "a" so that parabola y = ax² + 3x + 11 goes through the point (3, 2) is:

`\large \boxed{a=-2}`