Question

Which of the following statements about the sign of the quadratic expression \(\mathrm{E}=\mathrm{y}^{2}-12 \mathrm{y}+20\) is true? (1) \(\mathrm{E} \leq 0\) when \(2 \leq \mathrm{y} \leq 10\) and \(\mathrm{E}>0\) when \(\mathrm{y}<2\) or \(\mathrm{y}>10\) (2) \(\mathrm{E} \geq 0\) when \(2 \leq \mathrm{y} \leq 10\) and \(\mathrm{E}<0\) when \(\mathrm{y}<2\) or \(\mathrm{y}>10\) (3) \(\mathrm{E} \leq 0\) when \(-10 \leq \mathrm{y} \leq-2\) and \(\mathrm{E}>0\) when \(\mathrm{y}<-10\) or \(\mathrm{y}>-2\) (4) \(\mathrm{E} \geq-2\) and \(\mathrm{E}<0\) when \(\mathrm{y}<-10\) or \(\mathrm{y}>-2\)

Short answer

Question: Analyze the quadratic expression \(E=y^2-12y+20\), and choose the correct statement: (1) \(E \leq 0\) when \(2 \leq y \leq 10\) and \(E>0\) when \(y<2\) or \(y>10\) (2) \(E \geq 0\) when \(2 \leq y \leq 10\) and \(E<0\) when \(y<2\) or \(y>10\) (3) \(E \leq 0\) when \(6 \leq y \leq 20\) and \(E>0\) when \(y<6\) or \(y>20\) Answer: (1) \(E \leq 0\) when \(2 \leq y \leq 10\) and \(E>0\) when \(y<2\) or \(y>10\)

Step-by-step solution

1. Find the vertex of the parabola

In order to find the vertex of the parabola, we can use the vertex formula: Vertex (\(h,k\)), where \(h=\frac{-b}{2a}\) and \(k=f(h)\), given a quadratic in the form of \(f(y) = ay^2 + by + c\). In this case, \(a=1\), \(b=-12\), and \(c=20\). Hence, \(h =\frac{-(-12)}{2(1)}=6\) and \(k = f(6) = 6^2-12(6)+20 = -16\). So the vertex is at the point \((6,-16)\).

2. Identify the sign of the coefficient a

As the coefficient \(a=1\), which is positive, the parabola opens upwards and has its minimum value at the vertex, in this case, \(-16\).

3. Check the given statements

Now we have to check the given statements. (1) \(E \leq 0\) when \(2 \leq y \leq 10\) and \(E>0\) when \(y<2\) or \(y>10\) Since the vertex is \((6,-16)\) and the parabola opens upwards, the expression \(E\) will be non-positive in the interval \([2, 10]\) (i.e., \(E \leq 0\)). Let's check the other part of the statement. We know that for \(y<2\), we have the following signs: \(y^2>0\), and \(12y>0\). Thus, \(E=y^2-12y+20>0\). Similarly, for \(y>10\), \(y^2>0\), and \(12y>0\). Thus, \(E=y^2-12y+20>0\). So, statement (1) holds true. We conclude that statement (1) is the correct answer, and no need to check the other statements as there is only one correct answer.

Key concepts

Quadratic Expression Analysis

Understanding the sign of a quadratic expression is fundamental in graphing parabolas and problem-solving. A quadratic expression such as \(E=y^2-12y+20\) can take on positive or negative values depending on the value of \(y\). By analyzing the coefficients and completing the square or using the vertex form, one can determine where the expression is non-positive or non-negative.

For example, if the quadratic opens upwards (a positive 'a' coefficient), the expression will be at its minimum at the vertex. Beyond the vertex, on either side, the values of \(E\) become greater. In the case of our expression \(E\), since the coefficient of \(y^2\) is positive, we know the parabola opens upwards and thus, the range of values where \(E\) is non-positive is between the roots of the expression, if they exist.

Vertex of a Parabola

The vertex of a parabola is the point where the curve turns; it represents either the maximum or minimum point on the graph. For the parabola \(y=ax^2+bx+c\), the vertex \( (h,k) \) can be found using the formula \(h=-\frac{b}{2a}\) and \(k=f(h)\).

For the expression \(E\), the calculations show that the vertex is \( (6,-16) \). Since the 'a' coefficient is positive, this vertex is the lowest point on the parabola, indicating that \(E\) is at its minimum value of -16 when \(y=6\). Knowing the vertex provides significant insight into the behavior of the quadratic expression.

Coefficient Sign in Quadratic Equations

In quadratic equations, the sign of the leading coefficient 'a' is crucial as it dictates the direction in which the parabola opens. A positive 'a' means the arms of the parabola open upwards, and \(E\) will have its minimum value at the vertex. Conversely, a negative 'a' opens the parabola downwards, resulting in a maximum value at the vertex.

In our problem, the leading coefficient is 1 (positive), confirming that the parabola opens upwards and the vertex is a minimum point. This detail alone allows us to predict the change in sign of the quadratic expression as \(y\) increases or decreases.

Mathematical Problem-Solving

Effective mathematical problem-solving often requires analyzing relationships between variables and understanding the graphical representation of expressions. After finding the vertex of the parabola and understanding the direction in which it opens, one can apply this knowledge to determine the intervals where the quadratic expression will be non-positive or non-negative.

This process involved checking the given statements against our findings. The correct statement matched our analysis that the expression \(E\) is non-positive between \(2 \leq y \leq 10\) and positive otherwise. This problem-solving approach can be applied to a wide range of mathematical challenges involving quadratic expressions.