Question

Statement 1 Minimum value of the quadratic expression \(\left(x^{2}+x+4\right)\) is \(\frac{15}{4}\). and Statement 2 Minimum or maximum value of the quadratic expression \(\left(a x^{2}+b x+c\right)\) where \(a, b, c\) are real occurs at \(x=\frac{\alpha+\beta}{2}\) where \(\alpha\) and \(\beta\) are the roots of the equation \(a x^{2}+b x+c=0\).

Short answer

Answer: Yes, both Statement 1 and Statement 2 are true.

Step-by-step solution

Find the vertex and minimum value for Statement 1

For the quadratic expression \(x^2 + x + 4\), we have \(a = 1\), \(b = 1\), and \(c = 4\). To find the vertex, we use the formula \(x = \frac{-b}{2a}\). \(x = \frac{-1}{2(1)} = -\frac{1}{2}\) Now, we can plug this value of \(x\) back into the quadratic expression to find the minimum value: Minimum value = \(\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 4 = \frac{1}{4} - \frac{1}{2} + 4 = \frac{15}{4}\) Since the minimum value is indeed \(\frac{15}{4}\), Statement 1 is true.

Verify Statement 2 using Vieta's Formulas

Given that the roots of the quadratic equation \(ax^2 + bx + c = 0\) are \(\alpha\) and \(\beta\), according to Vieta's formulas: \(\alpha + \beta = \frac{-b}{a}\) and \(\alpha\beta = \frac{c}{a}\) Now, we are given that the minimum or maximum value occurs at \(x = \frac{\alpha+\beta}{2}\). We will show this by finding the vertex using the formula \(x = \frac{-b}{2a}\) and then prove that it is equal to \(\frac{\alpha+\beta}{2}\). Using the formula, we get: \(x = \frac{-b}{2a} = \frac{-(-\alpha-\beta)}{2} = \frac{\alpha + \beta}{2}\) So the vertex occurs at \(x = \frac{\alpha + \beta}{2}\), which means Statement 2 is true.

Key concepts

Minimum Value of Quadratics

Quadratic expressions often form a parabola when graphed. For a quadratic expression of the form \( ax^2 + bx + c \), the curve will open upwards if \( a > 0 \) and downwards if \( a < 0 \). The minimum value of the quadratic occurs at the vertex of the parabola. When the parabola opens upwards, the vertex represents the lowest point on the graph.

The quadratic expression in our case is \( x^2 + x + 4 \), where \( a = 1 \), \( b = 1 \), and \( c = 4 \). Using the vertex formula, \( x = \frac{-b}{2a} \), we can find that the vertex occurs at \( x = -\frac{1}{2} \).

Substituting \(-\frac{1}{2}\) back into the expression gives us the minimum value:

• \( \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 4 = \frac{1}{4} - \frac{1}{2} + 4 = \frac{15}{4} \)

This confirms that the minimum value of the quadratic expression \( x^2 + x + 4 \) is indeed \( \frac{15}{4} \).

Vertex of a Parabola

The vertex of a parabola is a crucial concept in understanding quadratic equations. It represents either the maximum or minimum point of the parabola, depending on its orientation.

For a quadratic equation in the standard form \( ax^2 + bx + c \), the vertex's x-coordinate can be found using the formula \( x = \frac{-b}{2a} \).

For our example, \( x^2 + x + 4 \), the values are \( a = 1 \), \( b = 1 \). Plugging these into the vertex formula:

• \( x = \frac{-1}{2 \times 1} = -\frac{1}{2} \)

This tells us that the vertex of the parabola is at \( x = -\frac{1}{2} \). When we calculate for \( y \), by substituting back into the expression, the vertex provides the corresponding y-value, giving the complete vertex point.

The vertex gives insight into the parabola's graph position and helps in determining the parabola's minimum or maximum value.

Vieta's Formulas

Vieta's Formulas provide a connection between the coefficients of a polynomial and sums and products of its roots. For a quadratic equation \( ax^2 + bx + c = 0 \), if \( \alpha \) and \( \beta \) are the roots, the formulas express the relationships as:

• \( \alpha + \beta = \frac{-b}{a} \)
• \( \alpha\beta = \frac{c}{a} \)

These formulas are not only useful in solving equations but also in verifying properties of quadratic equations.

In the context of our exercise, using Vieta's Formulas, the point where the minimum or maximum value occurs can be correlated with the vertex formula. We establish that the x-coordinate of the vertex \( x = \frac{-b}{2a} \) is equivalent to \( \frac{\alpha + \beta}{2} \). In essence, Vieta's shows that the calculated vertex coincides with the midpoint of the roots, linking algebraic and geometric interpretations.

Roots of a Quadratic Equation

The roots of a quadratic equation are the solutions to the equation \( ax^2 + bx + c = 0 \). These points are where the graph of the quadratic intersects the x-axis.

Quadratics can have two distinct real roots, one real root, or two complex roots depending on the discriminant \( \Delta = b^2 - 4ac \). A positive discriminant indicates two real roots, zero indicates one real root (a double root), and a negative discriminant indicates complex roots.

To find the roots, use the quadratic formula:

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

In relation to finding the vertex and minimum value, knowing the roots helps clarify the symmetry of the parabola and the properties of the quadratic. It's all interlinked, showcasing how various forms of expression provide different insights into the same mathematical object.