Problem 23 - Algebra and Trigonometry

Question

Do the graphs intersect in the given viewing rectangle? If they do, how many points of intersection are there? $$ y=-3 x^{2}+6 x-\frac{1}{2}, y=\sqrt{7-\frac{7}{12} x^{2}} ; \quad[-4,4] \text { by }[-1,3] $$

Step-by-Step Solution

Answer

The graphs intersect at two points within the specified viewing rectangle.

Step 1: Equate the Two Equations

To determine if the graphs intersect, set the equations equal: \(-3x^{2}+6x-\frac{1}{2} = \sqrt{7-\frac{7}{12} x^{2}}\). This step helps us find common values of \(x\) for which the functions are equal.

Step 2: Square Both Sides

Square both sides to eliminate the square root and make it easier to solve as an algebraic equation:\[\left(-3x^{2} + 6x - \frac{1}{2}\right)^{2} = 7 - \frac{7}{12}x^{2}\]

Step 3: Expand and Simplify

Expand the left side of the equation: \[\left(-3x^{2} + 6x - \frac{1}{2}\right)^{2} = 9x^{4} - 36x^{3} + 36x^{2} - 6x + \frac{1}{4} \]Set it equal to the expanded right side from Step 2 and bring all terms to one side.

Step 4: Solve the Polynomial Equation

The resulting equation is a high-degree polynomial equation, which may need estimation or graphing techniques for solutions. Use graphing software or analyze using quadratic or polynomial factoring techniques to verify solutions.

Step 5: Verify Within the Viewing Rectangle

Once solutions for \(x\) are found, substitute back to check that they fall within the given range \([-4,4]\) for the x-values and \([-1,3]\) for the y-values. This ensures all intersections are within the specified viewing rectangle.

Step 6: Conclusion on Points of Intersection

After plotting both equations on a graph or solving algebraically, determine the number of valid points within the given viewing rectangle. These are the points where both equations take the same y-value in the specified x-range.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that involve variables raised to the second power. The general form of a quadratic equation is given by \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not zero. These equations are called 'quadratic' because 'quad' refers to square, and the highest power of the variable is two. In our exercise, the function \( y = -3x^2 + 6x - \frac{1}{2} \) is a quadratic equation.
The graph of a quadratic equation is a parabola. It can open upwards or downwards depending on the sign of the coefficient \( a \). If \( a > 0 \), the parabola opens upwards, resembling a "U" shape. If \( a < 0 \), it opens downwards, resembling an inverted "U". In our case, since \( a = -3 \), the parabola opens downwards.
Finding solutions to a quadratic equation usually involves finding the values of \( x \) where the equation equals zero, also known as the roots or zeros of the equation. These solutions can be found through various methods, including factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)."}, {