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Math  /  Geometry

Question13. The vertex of the parabola y+16=(x−1)2y+16=(x-1)^{2} is located at (A) (−1,−16)(-1,-16) (B) (−1,16)(-1,16) (C) (1,−16)(1,-16) (D) (1,16)(1,16) (E) (16,1)(16,1)
14. Find the axis of symmetry of parabola

Studdy Solution

STEP 1

1. The equation of the parabola is given in vertex form: y+16=(x−1)2 y + 16 = (x - 1)^2 .
2. The vertex form of a parabola is y=a(x−h)2+k y = a(x - h)^2 + k , where (h,k)(h, k) is the vertex.
3. The axis of symmetry of a parabola in vertex form is x=h x = h .

STEP 2

1. Identify the vertex of the parabola.
2. Determine the axis of symmetry.

STEP 3

Identify the vertex of the parabola by comparing the given equation y+16=(x−1)2 y + 16 = (x - 1)^2 with the standard vertex form y=a(x−h)2+k y = a(x - h)^2 + k .
Rewriting the given equation in vertex form: y=(x−1)2−16 y = (x - 1)^2 - 16
From this, we can see that: - h=1 h = 1 - k=−16 k = -16
Thus, the vertex is (h,k)=(1,−16)(h, k) = (1, -16).

STEP 4

Determine the axis of symmetry using the vertex form of the parabola. The axis of symmetry is given by x=h x = h .
From the vertex (1,−16)(1, -16), the axis of symmetry is: x=1 x = 1
The vertex of the parabola is (1,−16)(1, -16), which corresponds to option (C).
The axis of symmetry of the parabola is x=1 x = 1 .

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