Question

write the equation of a parabola in vertex form that has a vertex at (3,14) and a y - intercept at (0,10). (1 point)

$y = -\frac{4}{9}(x - 3)^2+14$

$(x - 3)^2=-\frac{16}{9}(y - 14)$

$\frac{16}{9}(x - 3)=(y - 14)^2$

$y = \frac{4}{9}(x + 3)^2+14$

Explanation:

Step1: Recall vertex - form of parabola

The vertex - form of a parabola is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. Given the vertex $(h,k)=(3,14)$, the equation of the parabola is $y=a(x - 3)^2+14$.

Step2: Substitute the y - intercept

The y - intercept is at the point $(0,10)$. Substitute $x = 0$ and $y = 10$ into the equation $y=a(x - 3)^2+14$. We get $10=a(0 - 3)^2+14$.

Step3: Solve for a

First, simplify the equation: $10=a(-3)^2+14$, which is $10 = 9a+14$. Subtract 14 from both sides: $10-14=9a$, so $-4 = 9a$. Then $a=-\frac{4}{9}$.

Step4: Write the final equation

Substitute $a = -\frac{4}{9}$ back into the vertex - form equation. The equation of the parabola is $y=-\frac{4}{9}(x - 3)^2+14$.

Answer:

$y = -\frac{4}{9}(x - 3)^2+14$