Problem 13 - Beginning and Intermediate Algebra

Question

Find the vertex of each parabola. For each equation, decide whether the graph opens up, down, to the left, or to the right, and whether it is wider, narrower, or the same shape as the graph of \(y=x^{2}\). If it is a parabola with a vertical axis of symmetry, find the discriminant and use it to determine the number of \(x\) -intercepts. $$ f(x)=2 x^{2}+4 x+5 $$

Step-by-Step Solution

Answer

Vertex: (-1, 3), opens up, narrower than \(y = x^2\), no x-intercepts.

Step 1: Rewrite the Equation in Vertex Form

The given equation is in standard form: \(f(x) = 2x^2 + 4x + 5\). To find the vertex, rewrite it in vertex form, which is \(f(x) = a(x-h)^2 + k\).

Step 2: Complete the Square

Extract the coefficient of the quadratic term: \(f(x) = 2(x^2 + 2x) + 5\). Now complete the square inside the parenthesis: \(2(x^2 + 2x + 1 - 1) + 5 = 2((x+1)^2 - 1) + 5\).Simplify this to: \(2(x+1)^2 - 2 + 5 = 2(x+1)^2 + 3\). So, the vertex form is \(f(x) = 2(x + 1)^2 + 3\).

Step 3: Identify the Vertex

From the vertex form \(f(x) = 2(x + 1)^2 + 3\), identify the vertex \(h, k\), which is \(-1, 3\).

Step 4: Determine Direction and Shape

Since the coefficient \(a = 2\) is positive, the parabola opens upwards. The value of \(a = 2\) also means the parabola is narrower than the graph of \(y = x^2\) because \(|a| > 1\).

Step 5: Compute the Discriminant

For the quadratic equation \(ax^2 + bx + c\), the discriminant \(\text{D}\) is calculated as: \(\text{D} = b^2 - 4ac\). Substitute \(a = 2\), \(b = 4\), and \(c = 5\): \(\text{D} = 4^2 - 4 \cdot 2 \cdot 5 = 16 - 40 = -24\).

Step 6: Determine the Number of x-Intercepts

Since the discriminant \(\text{D} = -24\) is less than 0, the quadratic equation has no real roots, meaning there are no \(x\)-intercepts.

Key Concepts

Completing the Square

To find the vertex of a parabola given by a quadratic equation, one handy method is 'completing the square.' Start with the general quadratic expression. For instance, in our example, we have:

\(f(x) = 2x^2 + 4x + 5\).

To complete the square, first factor out the coefficient of the quadratic term (if necessary). This would transform the equation to:

\(f(x) = 2(x^2 + 2x) + 5\).

Next, find the term inside the parenthesis that makes a perfect square. Add and subtract this term inside the expression:

\(2[x^2 + 2x + 1 - 1] + 5\). Note, the term to add and subtract here is \(1\), since \(\frac{(2/2)^2}{2} = 1\).

This simplifies to:

\(2[(x+1)^2 - 1] + 5 = 2(x+1)^2 + 3\).

Thus, the vertex form of the original equation is \(f(x) = 2(x + 1)^2 + 3\). Completing the square allows us to easily identify and understand the properties of the parabola.

Quadratic Equations

A quadratic equation is a second-degree polynomial of the form:

\(ax^2 + bx + c = 0\).

It graphs as a parabola. The coefficient \(a\) influences the direction and the width:

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.
• The width of the parabola is determined by the absolute value of \(a\). If \(|a|>1\), the parabola is narrower than the standard \(y = x^2\) parabola. If \(|a|<1\), it is wider.

For example, in the equation \(2x^2 + 4x + 5\):

The term \(2x^2\) leads to a parabola that opens upwards and is narrower because the absolute value of \(2\) is more than \(1\). Finding the vertex helps visualize this by converting the quadratic equation into its vertex form. In our case, we identified the vertex as \((-1, 3)\). Therefore, the tip of the parabola is at point \((-1, 3)\).

Discriminant

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is denoted as \(D\) and given by the formula:

\[D = b^2 - 4ac\].

It helps determine the nature and number of roots of the quadratic equation:

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is one real root (or a repeated root).
• If \(D < 0\), there are no real roots, which indicates the parabola does not intersect the x-axis.

In our specific problem, substituting \(a = 2\), \(b = 4\), and \(c = 5\) into the discriminant formula:

\[D = 4^2 - 4 \times 2 \times 5 = 16 - 40 = -24\].

Given that \(D = -24\) (which is less than 0), it means the quadratic equation \(2x^2 + 4x + 5\) has no real roots. This means the parabola has no real x-intercepts.