Problem 13 - Algebra and Trigonometry

Question

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. $$ f(x)=x^{2}+4 x+3 $$

Step-by-Step Solution

Answer

The vertex is \((-2, -1)\). The x-intercepts are \(-1\) and \(-3\), with a y-intercept of \(3\). The graph is a parabola opening upwards.

Step 1: Identify Standard Form

The standard form of a quadratic function is given by \( f(x) = ax^2 + bx + c \). We need to confirm that the function \( f(x) = x^2 + 4x + 3 \) is in standard form. Here, \( a = 1 \), \( b = 4 \), and \( c = 3 \). It is already in standard form.

Step 2: Find the Vertex

To find the vertex of the quadratic function, we use the vertex formula \( x = -\frac{b}{2a} \). Substituting our values, \( a = 1 \) and \( b = 4 \):\[ x = -\frac{4}{2(1)} = -2 \]. To find the y-coordinate, substitute \( x = -2 \) back into the function: \( f(-2) = (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1 \). Therefore, the vertex is \((-2, -1)\).

Step 3: Find the x-Intercepts

To find the x-intercepts, solve the equation \( x^2 + 4x + 3 = 0 \). Factoring gives \( (x + 1)(x + 3) = 0 \). Thus, the x-intercepts are \( x = -1 \) and \( x = -3 \).

Step 4: Find the y-Intercept

To find the y-intercept, set \( x = 0 \) in the function \( f(x) = x^2 + 4x + 3 \). This results in \( f(0) = 0^2 + 4 \times 0 + 3 = 3 \). The y-intercept is \( 3 \).

Step 5: Sketch the Graph

Use the information gathered to sketch the graph. The parabola opens upwards because \( a = 1 \) is positive. Plot the vertex at \((-2, -1)\), the x-intercepts at \( (-1, 0) \) and \( (-3, 0) \), and the y-intercept at \( (0, 3) \). Draw a symmetric parabola opening upwards through these points.

Key Concepts

Vertex of a Quadratic Function

The vertex of a quadratic function is a crucial concept when graphing these types of functions because it represents the highest or lowest point on the parabola, depending on the direction it opens. Calculating the vertex can be done using the formula for the x-coordinate: \[ x = -\frac{b}{2a} \] where \( a \) and \( b \) are coefficients from the standard form equation of the quadratic function \( ax^2 + bx + c \). After identifying the x-coordinate, substitute it back into the original function to find the y-coordinate of the vertex.For example, with the function \( f(x) = x^2 + 4x + 3 \), the values are \( a = 1 \) and \( b = 4 \). Substituting into the formula gives \( x = -2 \). Plugging \( x = -2 \) back into the function calculates the y-value, thus the vertex is \((-2, -1)\).

X-Intercepts

Finding the x-intercepts of a quadratic function involves solving the quadratic equation \( ax^2 + bx + c = 0 \) for the values of \( x \). This is where the graph intersects the x-axis, meaning the y-value is zero at these points.For the function \( f(x) = x^2 + 4x + 3 \), solving the equation \( x^2 + 4x + 3 = 0 \) by factoring gives us the expression \((x + 1)(x + 3) = 0\). From this, the x-intercepts are \( x = -1 \) and \( x = -3 \). These points are where the parabola crosses the x-axis.

Y-Intercept

The y-intercept of a quadratic function is the point at which the parabola crosses the y-axis. To find this, we simply set \( x \) to zero and solve for \( f(x) \) in the quadratic function. For \( f(x) = x^2 + 4x + 3 \), substituting \( x = 0 \) gives \( f(0) = 0^2 + 4 \times 0 + 3 \). The resulting value for the y-intercept is \( 3 \). Therefore, the graph intersects the y-axis at the point \((0, 3)\).

Graphing Quadratic Functions

Graphing a quadratic function like \( f(x) = x^2 + 4x + 3 \) involves plotting key points such as the vertex and intercepts and then drawing a smooth curve.To begin, locate the vertex \((-2, -1)\) and plot it on the graph. Next, identify and plot the x-intercepts \((-1, 0)\) and \((-3, 0)\) and the y-intercept \((0, 3)\). The shape of the curve is a parabola, and because \( a = 1 \) (a positive value), the parabola opens upwards.- Plotting these points accurately on a coordinate grid helps in drawing a symmetrical parabolic curve.- The vertex serves as the axis of symmetry for the graph.This approach helps visualize the function more easily and understand how the equation translates to a visual graph.