Problem 45 - Algebra 1: Concepts and Skills

Question

Find the x-intercepts and the vertex of the graph of the function. Then sketch the graph of the function. \(y=(x+4)(x+3)\)

Step-by-Step Solution

Answer

The x-intercepts of \(y=(x+4)(x+3)\) are at \((-4,0)\) and \((-3,0)\) and the vertex of the function is at \((-7/2 ,-99/4)\).

Step 1: Find the x-intercepts

This can be calculated by setting \(y=(x+4)(x+3)\) equal to 0. We get: \[0=(x+4)(x+3)\] The solutions are \(x=-4\) and \(x=-3\), these are the x-intercepts.

Step 2: Find the vertex

The vertex of a quadratic function, \(y=ax^2+bx+c\), is found using \(-b/2a\) for the x-coordinate. The equation \(y=(x+4)(x+3)\) become \(y=x^2+7x+12\). So, by applying the vertex formula the x-coordinate of the vertex is \(-7/(2*1)=-7/2\). Plug this into the function to find the y-coordinate: \(y=(-7/2)^2+7*(-7/2)+12=-49/4-49/2+12=-49/4-98/4+48/4=-99/4\). Thus, the vertex of the function is \((-7/2 ,-99/4)\)

Step 3: Sketch the Graph

The x-intercepts are at \((-4, 0)\) and \((-3, 0)\), and the vertex is at \((-7/2 ,-99/4)\). Therefore, the parabola opens upwards. Given these points and the direction of opening, a sketch of the graph can be made.

Key Concepts

X-Intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. This occurs when the value of the function, \( y \), is zero. To find the x-intercepts, set the quadratic equation equal to zero and solve for \( x \).
For example, consider the function \( y = (x+4)(x+3) \). Set it to zero:

• \( 0 = (x+4)(x+3) \)

By solving this equation, you find the solutions that make the equation true:

• \( x = -4 \)
• \( x = -3 \)

These values represent the x-intercepts: the points \((-4, 0)\) and \((-3, 0)\). These intercepts provide key information about where the parabola will cross the x-axis, allowing us to better understand its graph.

Vertex of a Parabola

In the context of parabolas, the vertex represents the highest or lowest point on the graph. It is a crucial component because it gives you insight into the parabola's orientation and its most extreme value.
The vertex of a quadratic equation in standard form, \( y = ax^2 + bx + c \), can be found using the formula \( x = -\frac{b}{2a} \) to calculate the x-coordinate.
For the given equation \( y = (x+4)(x+3) \), which expands to \( y = x^2 + 7x + 12 \):

• The x-coordinate is \( x = -\frac{7}{2} \)

To find the y-coordinate, substitute back into the equation:

• \( y = \left( -\frac{7}{2} \right)^2 + 7 \left( -\frac{7}{2} \right) + 12 = -\frac{99}{4} \)

Therefore, the vertex of the parabola is at \( \left( -\frac{7}{2}, -\frac{99}{4} \right) \).
This point tells us that the parabola reaches its lowest point at this vertex since it opens upwards.

Graphing Parabolas

Graphing a parabola requires understanding its basic shape and critical points, such as x-intercepts and the vertex.
First, identify where the parabola crosses the x-axis by using the x-intercepts. For the function \( y = (x+4)(x+3) \), these points are \((-4, 0)\) and \((-3, 0)\).
The vertex, which is \( \left( -\frac{7}{2}, -\frac{99}{4} \right) \), offers the turning point of the parabola.

• The parabola opens upwards, as the coefficient of \( x^2 \) in the expanded form, \( y = x^2 + 7x + 12 \), is positive.
• You can plot the x-intercepts and the vertex to begin sketching the graph.
• Use the axis of symmetry, which runs vertically through the vertex, to ensure the parabola is symmetric.

By connecting these points smoothly, you create the characteristic U-shape of the parabola.
This helps visualize how the quadratic function behaves across different input values.