**Problem 33:** Find the points of intersection between the line $ x - y - 2 = 0 $ and the parabola $ y = 4 - x^2 $.**Solution:**To find the points of intersection, we need to solve the two equations simultaneously.Firstly, solve the linear equation for $ y $:\[ x - y - 2 = 0 \]\[ y = x - 2 \]Now, substitute $ y = x - 2 $ into the equation of the parabola:\[ y = 4 - x^2 \]\[ x - 2 = 4 - x^2 \]Rearrange to form a quadratic equation:\[ x^2 + x - 6 = 0 \]Next, solve the quadratic equation $ x^2 + x - 6 = 0 $ using the quadratic formula, factoring, or completing the square.Assuming factoring is possible:\[ (x + 3)(x - 2) = 0 \]Thus, the solutions are:\[ x = -3 \quad \text{or} \quad x = 2 \]Substitute these $ x $ values back into the linear equation to find the corresponding $ y $ values.For $ x = -3 $:\[ y = -3 - 2 = -5 \]Thus, one point of intersection is $ (-3, -5) $.For $ x = 2 $:\[ y = 2 - 2 = 0 \]Thus, the other point of intersection is $ (2, 0) $.**Conclusion:**The points of intersection between the line and the parabola are $ (-3, -5) $ and $ (2, 0) $.

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Answered: 33) Find the points of intersection…

Math

Geometry

33) Find the points of intersection between the line x - y - 2 = 0 and the parabola y = 4 - x 2.

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

Question

**Problem 33:** Find the points of intersection between the line $ x - y - 2 = 0 $ and the parabola $ y = 4 - x^2 $.

**Solution:**

To find the points of intersection, we need to solve the two equations simultaneously.

Firstly, solve the linear equation for $ y $:
\[ x - y - 2 = 0 \]
\[ y = x - 2 \]

Now, substitute $ y = x - 2 $ into the equation of the parabola:
\[ y = 4 - x^2 \]
\[ x - 2 = 4 - x^2 \]

Rearrange to form a quadratic equation:
\[ x^2 + x - 6 = 0 \]

Next, solve the quadratic equation $ x^2 + x - 6 = 0 $ using the quadratic formula, factoring, or completing the square.

Assuming factoring is possible:
\[ (x + 3)(x - 2) = 0 \]

Thus, the solutions are:
\[ x = -3 \quad \text{or} \quad x = 2 \]

Substitute these $ x $ values back into the linear equation to find the corresponding $ y $ values.

For $ x = -3 $:
\[ y = -3 - 2 = -5 \]
Thus, one point of intersection is $ (-3, -5) $.

For $ x = 2 $:
\[ y = 2 - 2 = 0 \]
Thus, the other point of intersection is $ (2, 0) $.

**Conclusion:**

The points of intersection between the line and the parabola are $ (-3, -5) $ and $ (2, 0) $.

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