The problem states that $y$ varies directly as $x^2$. We are given a table of corresponding values for $x$ and $y$. We need to find: a. The constant of variation $k$. b. The equation connecting $y$ and $x$. c. The value of $a$.

1. Problem Description

The problem states that $y$ varies directly as $x^2$. We are given a table of corresponding values for $x$ and $y$. We need to find:

a. The constant of variation $k$.

b. The equation connecting $y$ and $x$.

c. The value of $a$.

2. Solution Steps

a. Since $y$ varies directly as $x^2$, we have the relationship

$y = kx^2$

We can use the values $x=4$ and $y=96$ to find $k$:

$96 = k(4^2)$

$96 = 16k$

$k = \frac{96}{16}$

$k = 6$

b. The equation connecting $y$ and $x$ is given by $y = kx^2$. We found that $k=6$, so the equation is

$y = 6x^2$

c. We are given that when $x=a$, $y=24$. We can substitute these values into the equation we found in part (b) to find $a$.

$24 = 6a^2$

$a^2 = \frac{24}{6}$

$a^2 = 4$

$a = \pm\sqrt{4}$

$a = \pm 2$

3. Final Answer

a. The constant of variation, $k = 6$.

b. The equation connecting $y$ and $x$ is $y = 6x^2$.

c. The values of $a$ are $a = 2$ and $a = -2$.