Problem 10 - Basic Technical Mathematics with Calculus

Question

Sketch the graph of each parabola by using the vertex, the \(y\) -intercept, and the \(x\) -intercepts. Check the graph using \(a\) calculator. \(y=x^{2}+3 x\)

Step-by-Step Solution

Answer

The vertex is at \((-1.5, -2.25)\), the \(y\)-intercept is \((0,0)\), and the \(x\)-intercepts are \((0,0)\) and \((-3,0)\). The parabola opens upward.

Step 1: Identify the Standard Form

The standard form of a parabola is given by \( y = ax^2 + bx + c \). For the equation \( y = x^2 + 3x \), we have \( a = 1 \), \( b = 3 \), and \( c = 0 \).

Step 2: Find the Vertex

The vertex of a parabola in standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = 3 \) into the formula gives \( x = -\frac{3}{2} = -1.5 \). Now substitute this back into the equation to find \( y \):\( y = (-1.5)^2 + 3(-1.5) = 2.25 - 4.5 = -2.25 \).So, the vertex is \((-1.5, -2.25)\).

Step 3: Find the Y-Intercept

The \( y \)-intercept is found by setting \( x = 0 \). Substitute \( x = 0 \) into the equation:\( y = 0^2 + 3 \cdot 0 = 0 \).So, the \( y \)-intercept is \((0, 0)\).

Step 4: Find the X-Intercepts

The \( x \)-intercepts are found by setting \( y = 0 \). Solve the equation:\( 0 = x^2 + 3x \).Factor the equation:\( x(x + 3) = 0 \).Setting each factor to zero gives the solutions \( x = 0 \) and \( x = -3 \).So, the \( x \)-intercepts are \((0, 0)\) and \((-3, 0)\).

Step 5: Sketch the Graph

Plot the vertex \((-1.5, -2.25)\), \( y \)-intercept \((0, 0)\), and \( x \)-intercepts \((0, 0)\) and \((-3, 0)\) on a coordinate grid. Draw a parabola that passes through these points. The parabola opens upward as the coefficient of \( x^2 \) is positive.

Step 6: Check with a Calculator

Use a graphing calculator to plot the function \( y = x^2 + 3x \). Check to ensure that the vertex, \( y \)-intercept, and \( x \)-intercepts match the derived points from the previous steps.

Key Concepts

Vertex of a Parabola

One of the essential features when sketching a parabola is finding its vertex. The vertex is the point where the parabola changes direction, acting as either the lowest or highest point depending on whether the parabola opens upward or downward.
In the standard form of a quadratic equation, which is written as \( y = ax^2 + bx + c \), the vertex can be found using the formula \( x = -\frac{b}{2a} \).
For the parabola given by \( y = x^2 + 3x \), with \( a = 1 \) and \( b = 3 \), substituting these values into the formula gives us \( x = -\frac{3}{2} \). This simplifies to \( -1.5 \).

• The \( y \)-coordinate of the vertex can be found by substituting \( x = -1.5 \) back into the equation: \( y = (-1.5)^2 + 3(-1.5) = 2.25 - 4.5 = -2.25 \).
• Thus, the vertex is located at \((-1.5, -2.25)\).

Finding the vertex helps in understanding the symmetry of the parabola and is crucial for graphing.

Y-Intercept

To find the point where the parabola crosses the \( y \)-axis, you need to determine the \( y \)-intercept. It's where the value of \( x \) is equal to zero.
Using the quadratic equation \( y = x^2 + 3x \), set \( x = 0 \) and solve for \( y \).
This leads to \( y = 0^2 + 3 \times 0 = 0 \).

• The \( y \)-intercept of this parabola is thus \((0, 0)\).

Not only is the \( y \)-intercept useful when sketching the graph of the parabola, but it also gives you a clear point through which your graph must pass.

X-Intercepts

The \( x \)-intercepts provide critical points where the parabola intersects the \( x \)-axis. These occur where the value of \( y \) is zero.
Given the equation \( y = x^2 + 3x \), set \( y = 0 \) and solve for \( x \): \[ 0 = x^2 + 3x \]
This can be factored into \( x(x + 3) = 0 \).

• This gives the solutions \( x = 0 \) and \( x = -3 \).
• Therefore, the \( x \)-intercepts are at \((0, 0)\) and \((-3, 0)\).

Finding these intercepts provides two additional points necessary for sketching the curve accurately.

Standard Form of a Quadratic Equation

Understanding the standard form of a quadratic equation is foundational to graphing any parabola. This form is expressed as \( y = ax^2 + bx + c \), where:

• \( a \) determines the direction of the parabola's opening. If \( a \) is positive, the parabola opens upward; if negative, it opens downward.
• \( b \) influences the position of the vertex along the \( x \)-axis.
• \( c \) represents the \( y \)-intercept, providing a clear point of contact with the \( y \)-axis.

In the given equation \( y = x^2 + 3x \), interpreting these coefficients gives insights into the shape and orientation of the parabola. With \( a = 1 \), it is clear that the parabola opens upwards. Recognizing this form is vital for identifying the needed steps to determine other key features such as the vertex and intercepts.