Certainly! When we say that [tex]\( y \)[/tex] varies directly as [tex]\( x^2 \)[/tex], it means there is a constant [tex]\( k \)[/tex] such that:
[tex]\[ y = k \cdot x^2 \][/tex]
Given the values [tex]\( y = 16 \)[/tex] and [tex]\( x = -4 \)[/tex], we can use these to find the value of [tex]\( k \)[/tex].
1. Start with the direct variation equation:
[tex]\[ y = k \cdot x^2 \][/tex]
2. Substitute the given values [tex]\( y = 16 \)[/tex] and [tex]\( x = -4 \)[/tex] into the equation:
[tex]\[ 16 = k \cdot (-4)^2 \][/tex]
3. Calculate [tex]\( (-4)^2 \)[/tex]:
[tex]\[ (-4)^2 = 16 \][/tex]
4. Substitute this back into the equation:
[tex]\[ 16 = k \cdot 16 \][/tex]
5. Solve for [tex]\( k \)[/tex] by dividing both sides of the equation by 16:
[tex]\[ k = \frac{16}{16} \][/tex]
6. Simplify the fraction:
[tex]\[ k = 1 \][/tex]
Therefore, the value of the constant [tex]\( k \)[/tex] is [tex]\( 1.0 \)[/tex].
