Problem 13 - Algebra and Trigonometry

Question

\(5-18=\) A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x-\) and \(y\) -intercept(s). (c) Sketch its graph. $$ f(x)=2 x^{2}+4 x+3 $$

Step-by-Step Solution

Answer

Vertex: \((-1,1)\), no x-intercepts, y-intercept: \((0,3)\).

Step 1: Identify the given function

The given quadratic function is already provided in standard form as \( f(x) = 2x^2 + 4x + 3 \). Standard form of a quadratic function is generally expressed as \( ax^2 + bx + c \).

Step 2: Find the vertex

To find the vertex, use the formula for the vertex \( h = -\frac{b}{2a} \) where \( a = 2 \) and \( b = 4 \). So, \( h = -\frac{4}{2 \times 2} = -1 \). Substitute \( x = -1 \) back into the function to find the \( k \): \( k = 2(-1)^2 + 4(-1) + 3 = 1 \). Thus, the vertex is \((-1, 1)\).

Step 3: Find the x-intercepts

To find the x-intercepts, set \( f(x) = 0 \): \( 2x^2 + 4x + 3 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting, \( a = 2, b = 4, c = 3 \):\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \times 2 \times 3}}{2 \times 2} = \frac{-4 \pm \sqrt{-8}}{4} \]This gives no real roots since the discriminant is negative, so there are no x-intercepts.

Step 4: Find the y-intercept

The y-intercept is found by setting \( x = 0 \) in the function: \( f(0) = 2(0)^2 + 4(0) + 3 = 3 \). So, the y-intercept is \((0, 3)\).

Step 5: Sketch the graph

With the vertex at \((-1,1)\) and the y-intercept at \((0,3)\), sketch a parabola opening upwards because the coefficient of \( x^2 \), \( a = 2 \), is positive. Note the parabola does not cross the x-axis since there are no real “x” intercepts.

Key Concepts

Standard Form of Quadratic Function

A quadratic function in its standard form is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). This form is used to identify the basic elements of a quadratic equation, such as its shape and direction. The equation \( f(x) = 2x^2 + 4x + 3 \) is an example of a quadratic function in standard form.

• The coefficient \( a = 2 \) dictates the parabolic curve opens upwards because \( a > 0 \).
• The coefficient \( b = 4 \) influences the slope of the graph at any point.
• The constant term \( c = 3 \) is the y-intercept but more on this later.

Recognizing this standard form makes it easier to perform various operations, like finding the vertex or intercepts. Just remember: keeping an eye on the values of \( a \), \( b \), and \( c \) helps in sketching and understanding the parabola's behavior.

Vertex of Quadratic Function

The vertex of a quadratic function is the highest or lowest point of its parabola, depending on whether the parabola opens upwards or downwards. It serves as a significant point on the graph because it indicates the maximum or minimum value of the function.To find the vertex of \( f(x) = 2x^2 + 4x + 3 \), we use the vertex formula \( h = -\frac{b}{2a} \):

• Given \( a = 2 \) and \( b = 4 \).
• Calculate \( h = -\frac{4}{2 \times 2} = -1 \).

Once the \( x \)-coordinate (\( h \)) is found, substitute it back into the function to find the \( y \)-coordinate (\( k \)):

• \( k = 2(-1)^2 + 4(-1) + 3 = 1 \).

Therefore, the vertex is at \((-1, 1)\). This vertex indicates that the parabola reaches its minimum value at \(-1\), with a value of \( f(x) = 1 \). Knowing the vertex is crucial for understanding where the graph turns, providing insights into the function's behavior.

Intercepts of Quadratic Function

Intercepts are where the graph intersects the coordinate axes. They are essential for sketching the graph accurately and understanding the function's real-world context. There are two types of intercepts: x-intercepts and y-intercepts.**X-Intercepts:**An x-intercept occurs where the graph crosses the x-axis, where \( f(x) = 0 \). For \( f(x) = 2x^2 + 4x + 3 \), solving for x-intercepts involves setting:\[ 2x^2 + 4x + 3 = 0 \]Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute \( a = 2 \), \( b = 4 \), \( c = 3 \):

• The discriminant, \( b^2 - 4ac = 4^2 - 4 \times 2 \times 3 = -8 \), is negative.

Since the discriminant is negative, there are no real x-intercepts, meaning the parabola does not cross the x-axis.**Y-Intercept:**The y-intercept is found by setting \( x = 0 \) in the function:

• \( f(0) = 2(0)^2 + 4(0) + 3 = 3 \).

Thus, the y-intercept is \((0, 3)\), meaning the graph crosses the y-axis at this point. Knowing the intercepts is pivotal as it anchors the graph and outlines its position relative to the axes.