Question

The parabola \(y=4-x^{2}\) has vertex \(P\). It intersects the \(x\)-axis at \(A\) and \(B\). If the parabola is translated from its initial position to a new position by moving its vertex along the line \(y=x+4\), so that it intersects the \(x\)-axis at \(B\) and \(C\), then the abscissa of \(C\) will be:

• A. 12
• B. 8
• C. 6
• D. $\frac{7}{3}$

Hint:

Parabolas are perfectly symmetrical! Since the original parabola has a width of 4 units (from $-2$ to $+2$), the translated parabola must also have a width of exactly 4 units on either side of its new vertex axis $x = 5$. Since one intercept is at $B(2)$, the other intercept $C$ must be sitting symmetrically at $5 + 3 = 8$ units!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Translation of a parabola preserves its shape (curvature) but changes its vertex coordinates. The general form for a vertically opening parabola with vertex \( (h, k) \) is \( y - k = a(x - h)^2 \).
Step 2: Key Formula or Approach:
1. Initial parabola: \( y = -1(x - 0)^2 + 4 \). Vertex \( P(0, 4) \).
2. New vertex \( V(h, k) \) lies on \( y = x + 4 \), so \( k = h + 4 \).
3. New parabola: \( y = -(x - h)^2 + h + 4 \).
Step 3: Detailed Explanation:
The initial parabola intersects the x-axis at \( B(2, 0) \) and \( A(-2, 0) \).
The translated parabola also passes through \( B(2, 0) \).
Substitute \( (2, 0) \) into the new equation:
\[ 0 = -(2 - h)^2 + h + 4 \]
\[ (2 - h)^2 = h + 4 \]
\[ 4 - 4h + h^2 = h + 4 \implies h^2 - 5h = 0 \]
This gives \( h = 0 \) (original position) or \( h = 5 \).
The new vertex is at \( (5, 9) \).
The equation for the new parabola is \( y = -(x - 5)^2 + 9 \).
To find intersection points with x-axis, set \( y = 0 \):
\[ (x - 5)^2 = 9 \implies x - 5 = \pm 3 \]
\[ x = 5 + 3 = 8 \text{ and } x = 5 - 3 = 2 \].
The x-intercepts are 2 (Point B) and 8 (Point C).
The abscissa of C is 8.
Step 4: Final Answer:
The abscissa of C is 8.