Problem 41 - 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. \(y=(x+5)(x+3)\)

Step-by-Step Solution

Answer

The x-intercepts are \(x=-5, -3\) and the vertex is \(-4, -1\). The graph is a parabola opening upwards with its minimum point at \(-4, -1\) and passing through the points \((-5,0)\) and \((-3,0)\).

Step 1: Find the x-intercepts

Set the function equal to zero and solve for \(x\).\(0 = (x+5)(x+3)\)This will give us two solutions (the x-intercepts): \(x=-5, -3\).

Step 2: Find the vertex of the function

First, write the equation in the form \(ax^2 + bx + c\), i.e., \(y= x^2 +8x +15\). The x-coordinate of the vertex is \(-b/2a\). For this equation, \(a=1\) and \(b=8\), so \(-b/2a = -8/2=-4\).Then, substitute \(-4\) into \(y=x^2 +8x +15\), and we will find \(y=-1\). Therefore, the vertex of the function is at \((-4, -1)\).

Step 3: Sketch the graph

The x-intercepts are at \(x=-5, -3\), the vertex is at \((-4, -1)\), and as \(y=x^2 +8x +15\) is a parabola opening upwards, the sketch will be a curve with a minimum point at the vertex, passing through the x-intercepts.

Key Concepts

x-intercepts of a parabola

Understanding the x-intercepts of a parabola is essential in graphing quadratic functions. The x-intercepts, also known as the roots or zeros of the function, are points where the graph crosses the x-axis. These intercepts reveal important features of the parabola and are found by setting the quadratic function equal to zero.

To find these crucial points when given a quadratic function in factored form, such as in our exercise \(y=(x+5)(x+3)\), you solve for \(x\) when \(y=0\). This leads to \(x+5=0\) or \(x+3=0\), which further simplifies to \(x=-5\) and \(x=-3\). These are the x-intercepts that are part of the graph's structure and must be reflected when graphing the function.

vertex of a parabola

The vertex of a parabola represents the peak or trough of its curve, which can be either the highest or lowest point on the graph, depending on the direction in which the parabola opens. For the function \(y=x^2+8x+15\), we're dealing with an upward-opening parabola, so the vertex will be the lowest point.

To find the vertex, you can use the formula \(h=-b/(2a)\) for the x-coordinate and then substitute \(h\) back into the equation to get the y-coordinate, \(k\). With \(a=1\) and \(b=8\) as seen from the standard form of a quadratic \(ax^2+bx+c\), the vertex's x-coordinate is \(h=-8/2=-4\). Substituting \(x=-4\) into the quadratic gives us \(y=(-4)^2+8(-4)+15=-1\), leading to the vertex's coordinates \( (-4, -1) \). The vertex is a fundamental point used in the graphing process.

sketching parabolas

Sketching parabolas is a graphical way of visualizing the behavior of quadratic functions. Once the x-intercepts and the vertex of the parabola have been determined, as we did in our exercise with intercepts at \(x=-5\) and \(x=-3\), and the vertex at \( (-4, -1)\), sketching can begin.

Start by plotting the x-intercepts and vertex on the coordinate plane. Since our given function is an upward-opening parabola, the vertex will be at the bottom of the curve. Next, remember that parabolas are symmetrical about a vertical line that passes through the vertex, known as the axis of symmetry. In our case, this line is \(x=-4\). Sketch the parabola by drawing a curve through the x-intercepts that bows upward, making sure to reflect the parabola's symmetry across the axis. Adding a few more points by choosing x-values and computing the corresponding y-values can help in making your sketch more accurate and comprehensive.

quadratic functions

Quadratic functions are mathematical expressions of the form \(y=ax^2+bx+c\) where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). They represent parabolas when graphed on a coordinate plane. These functions are characterized by their curved graphs, which can open upward or downward depending on the sign of \(a\); positive for upward and negative for downward.

The graph of a quadratic is determined by its x-intercepts, vertex, and direction of opening. For the function \(y=(x+5)(x+3)\) which is equivalent to \(y=x^2+8x+15\), \(a=1\), \(b=8\), and \(c=15\). This tells us that the parabola opens upwards since \(a>0\) and has a vertex representing the lowest point on the graph. These features are pivotal for understanding the function's behavior and for predicting values for \(y\) based on input values for \(x\).