Question

fill in the blank 14 points
the quadratic function ( y = - \frac { 1 } { 16 } x ^ { 2 } + 1 ) has a vertex at ( ( h, k ) ) and focal length of ( c ^ { prime } ).
what are the values for ( h, k ), and ( c ^ { prime } )?
( h = ) type your answer...
( k = ) type your answer...
( c ^ { prime } = ) type your answer...
the focal point for ( y = - \frac { 1 } { 16 } x ^ { 2 } + 1 ) is at ( ( 0, p ) ). what is the value of ( p )?
( p = ) type your answer...

Explanation:

Step1: Recall the vertex form of a parabola

The general form of a parabola is \(y = a(x - h)^2 + k\). For the given function \(y=-\frac{1}{16}x^{2}+1\), we can rewrite it as \(y =-\frac{1}{16}(x - 0)^2+1\).

Step2: Identify \(h\) and \(k\)

Comparing with \(y = a(x - h)^2 + k\), we get \(h = 0\) and \(k = 1\).

Step3: Recall the formula for the focal length \(c\)

For a parabola \(y=ax^{2}+bx + c\) (or \(y=a(x - h)^2 + k\)), the focal length \(c=\frac{1}{4|a|}\). Here \(a =-\frac{1}{16}\), so \(|a|=\frac{1}{16}\). Then \(c=\frac{1}{4\times\frac{1}{16}}\).

$$ c = 4 $$

Step4: Find the \(y -\) coordinate of the focal point \(p\)

The focal point of a parabola \(y=a(x - h)^2 + k\) is \((h,k + c)\) (since for \(y=a(x - h)^2 + k\), the parabola opens up if \(a>0\) and down if \(a < 0\); here \(a=-\frac{1}{16}<0\) and \(c = 4\)). Substituting \(h = 0,k = 1,c = 4\) into \(p=k - c\) (because the parabola \(y=-\frac{1}{16}x^{2}+1\) opens downwards), we get \(p=1-4=- 3\).

Answer:

\(h = 0\)
\(k = 1\)
\(c = 4\)
\(p=-3\)