Question

If \(y=\min \left(x^{2}+2,6-3 x\right)\), then the greatest value of \(y\) for \(x>0\) (a) 1 (b) 2 (c) 3 (d) none of these

Short answer

Answer: (c) 3

Step-by-step solution

Find the Intersection Points

Since we want to find where the two functions intersect on the \(y-\)axis, we need to find the point where \(x^2 + 2 = 6 - 3x\). To do this, we will solve the equation: \(x^2 + 2 = 6 - 3x\)

Solve the Equation

Rearrange the equation by moving all the terms to the left side: \(x^2 + 3x - 4 = 0\) Now apply the quadratic formula to solve for \(x\). The quadratic formula is: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) For our equation, \(a=1\), \(b=3\), and \(c=-4\). Plugging these values into the quadratic formula, we get: \(x=\frac{-3\pm\sqrt{3^2-4(1)(-4)}}{2(1)}\) Evaluating the expression, we get: \(x=\frac{-3\pm\sqrt{25}}{2}\) So, we have two possible values for \(x\): \(x_1=\frac{-3+5}{2}=1\) \(x_2=\frac{-3-5}{2}=-4\) Since the problem states that we're only considering \(x > 0\), we ignore \(x_2=-4\) and use only \(x_1 = 1\).

Determine y values.

Now that we know the intersection point lies at \(x = 1\), we can plug this value back into either \(y = x^2 + 2\) or \(y = 6 - 3x\) to find the corresponding \(y\) value. Plugging \(x = 1\) into \(y = x^2 + 2\): \(y = (1)^2 + 2 = 3\) Thus, the intersection takes place at \((1, 3)\).

Find the maximum value of y for x > 0.

Now that we have the intersection point, we can find the maximum value of \(y\) for \(x > 0\). For \(x > 1\), the function \(y = 6 - 3x\) is less than \(y = x^2 + 2\), so we analyze only the quadratic function to the right of the intersection point. Since \(y = x^2 + 2\) is a parabola that opens upwards, the value of \(y\) will keep increasing as \(x\) increases for \(x > 1\). Therefore, we've found the maximum possible value of \(y\) at the intersection point \(y = 3\). So, the answer is (c) 3.

Key concepts

Quadratic Formula

The quadratic formula is a reliable tool to solve quadratic equations of the form \(ax^2 + bx + c = 0\). It allows us to find the roots or solutions of these equations, which are the values of \(x\) that make the equation true. The formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

To apply the quadratic formula, identify coefficients \(a\), \(b\), and \(c\) from your equation. Plug these into the formula to solve for \(x\). There are often two solutions, represented by the \(\pm\) in the formula, meaning you will perform the calculation twice: once with a plus sign and once with a minus sign.
In our problem, the equation \(x^2 + 3x - 4 = 0\) was solved using the quadratic formula, with \(a = 1\), \(b = 3\), and \(c = -4\). The solutions obtained were \(x = 1\) and \(x = -4\). Since we're considering \(x > 0\), we focus only on \(x = 1\). This example demonstrates the formula's usefulness in finding intersection points for comparing expressions.

Graphing Parabolas

A parabola is a symmetrical curve on the graph, and it can open either upwards or downwards. When dealing with quadratic equations in the form \(y = ax^2 + bx + c\), the graph will be a parabola.
Understanding the graph's shape and direction is crucial. If \(a > 0\), the parabola opens upwards, and if \(a < 0\), it opens downwards. The vertex, which is the highest or lowest point on a parabola, plays a significant role in determining its characteristics.

• The vertex can be found using the formula for \(x\): \(x = -\frac{b}{2a}\).
• Substitute this \(x\) back into the equation to find the \(y\) value.

In our exercise, the quadratic function \(y = x^2 + 2\) is analyzed. It’s a standard upward opening parabola. As \(x\) gets larger, so does \(y\), illustrating that the maximum \(y\)-value at the intersection point \((1, 3)\) is considered for \(x > 0\). Graphing helps visualize these interpretations of the problem.

Maxima and Minima

Maxima and minima refer to the highest and lowest points, respectively, that a function can reach. These are critical in analyzing the behavior of functions, especially parabolas, because they can indicate optimal or boundary points.
For the quadratic function \(y = ax^2 + bx + c\):

• The maximum or minimum depends on the sign of \(a\).
• If \(a > 0\), there's a minimum value (vertex is the lowest point).
• If \(a < 0\), there's a maximum value (vertex is the highest point).

In this problem, because we looked at \(y = x^2 + 2\), our function had a minimum, as it opens upwards due to \(a = 1\). The intersection at \((1, 3)\) provided the greatest \(y\)-value for \(x > 0\), since beyond \(x = 1\), \(y\) of \(6 - 3x\) reduces. Therefore, the solution to the problem was influenced by locating this point of maximum \(y\) value in the context given.