13. The curve $$ y=-x^{2} $$ is shifted so that its axis of symmetry is the line $$ x=1 $$ and its orthogonal axis is $$ y=3 $$. a. Write down the equation of the new curve. b. Find the coordinates of the points where this new curve cuts the $$ x $$ and $$ y $$ axes. c. Sketch the curve

Question

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the question.

### Part a: Write down the equation of the new curve.

The original curve is given by the equation:

$$
y = -x^2
$$

This curve has its axis of symmetry along the line $$ x = 0 $$. We need to shift this curve so that its axis of symmetry is the line $$ x = 1 $$ and its orthogonal axis is $$ y = 3 $$.

1. **Shift the axis of symmetry**: To shift the axis of symmetry from $$ x = 0 $$ to $$ x = 1 $$, we replace $$ x $$ with $$ x - 1 $$:

$$
y = -(x - 1)^2
$$

2. **Shift the orthogonal axis**: To shift the curve vertically so that it is centered around $$ y = 3 $$, we add 3 to the equation:

$$
y = -(x - 1)^2 + 3
$$

Thus, the equation of the new curve is:

$$
y = - (x - 1)^2 + 3
$$

### Part b: Find the coordinates of the points where this new curve cuts the $$ x $$ and $$ y $$ axes.

1. **Finding the $$ x $$-intercepts**: To find where the curve cuts the $$ x $$-axis, we set $$ y = 0 $$:

$$
0 = - (x - 1)^2 + 3
$$

Rearranging gives:

$$
(x - 1)^2 = 3
$$

Taking the square root of both sides:

$$
x - 1 = \pm \sqrt{3}
$$

Thus, we have:

$$
x = 1 + \sqrt{3} \quad \text{and} \quad x = 1 - \sqrt{3}
$$

So the $$ x $$-intercepts are:

$$
(1 + \sqrt{3}, 0) \quad \text{and} \quad (1 - \sqrt{3}, 0)
$$

2. **Finding the $$ y $$-intercept**: To find where the curve cuts the $$ y $$-axis, we set $$ x = 0 $$:

$$
y = - (0 - 1)^2 + 3 = -1 + 3 = 2
$$

So the $$ y $$-intercept is:

$$
(0, 2)
$$

### Part c: Sketch the curve

To sketch the curve, we note the following:

- The vertex of the parabola is at the point $$ (1, 3) $$.
- The parabola opens downwards (since the coefficient of the squared term is negative).
- The $$ x $$-intercepts are at $$ (1 + \sqrt{3}, 0) $$ and $$ (1 - \sqrt{3}, 0) $$.
- The $$ y $$-intercept is at $$ (0, 2) $$.

Here is a rough sketch of the curve:

```
   y
   |
  4|          *
  3|         * *
  2|        *   *
  1|       *     *
  0|------*-------*-------*------ x
   |     (1-√3)  1   (1+√3)
```

The vertex is at $$ (1, 3) $$, and the curve intersects the $$ y $$-axis at $$ (0, 2) $$ and the $$ x $$-axis at the two points calculated above. The parabola is symmetric about the line $$ x = 1 $$.
a. The equation of the new curve is $$ y = - (x - 1)^2 + 3 $$. b. The $$ x $$-intercepts are $$ (1 + \sqrt{3}, 0) $$ and $$ (1 - \sqrt{3}, 0) $$, and the $$ y $$-intercept is $$ (0, 2) $$. c. The vertex is at $$ (1, 3) $$, and the curve intersects the $$ y $$-axis at $$ (0, 2) $$ and the $$ x $$-axis at $$ (1 + \sqrt{3}, 0) $$ and $$ (1 - \sqrt{3}, 0) $$.

The curve is shifted so that its axis of symmetry is the line and
its orthogonal axis is .
a. Write down the equation of the new curve.
b. Find the coordinates of the points where this new curve cuts the and
axes.
c. Sketch the curve

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Pre Calculus Questions

The curve is shifted so that its axis of symmetry is the line and its orthogonal axis is . a. Write down the equation of the new curve. b. Find the coordinates of the points where this new curve cuts the and axes. c. Sketch the curve