Question

Graph the function \(3 \mathrm{x}^{2}+5 \mathrm{x}-7\).

Short answer

The function \(3x^2 + 5x -7\) graphs a parabola that opens upwards with vertex \(\left(-\frac{5}{6}, -\frac{85}{12}\right)\), x-intercepts \(\frac{-5 + \sqrt{109}}{6}\) and \(\frac{-5 - \sqrt{109}}{6}\), and y-intercept \((0, -7)\). The graph is symmetric about axis \(x = -\frac{5}{6}\).

Step-by-step solution

1. Determine the direction of the parabola

The function is given in the form \(f(x) = ax^2 + bx + c\), where \(a = 3\), \(b = 5\), and \(c = -7\). Since \(a\) is positive, the parabola opens upwards.

2. Calculate the vertex by completing the square

To find the vertex, we need the coordinates \((h, k)\). We can find them by completing the square: \[f(x) = 3x^2 + 5x - 7\] First, factor out the leading coefficient, \(a\): \[f(x) = 3(x^2 + \frac{5}{3}x) - 7\] Now, complete the square within the parenthesis: \[f(x) = 3\left(x^2 + \frac{5}{3}x + \left(\frac{5}{6}\right)^2 - \left(\frac{5}{6}\right)^2\right) - 7\] \[f(x) = 3\left(\left(x + \frac{5}{6}\right)^2 - \frac{25}{36}\right) - 7\] Now we can determine the vertex coordinates, \((h, k)\): \[h = -\frac{5}{6}\] and \[k = 3\left(-\frac{25}{36}\right) - 7 = -\frac{85}{12}\] So, the vertex of the parabola is at the point \(\left(-\frac{5}{6}, -\frac{85}{12}\right)\).

3. Calculate the x- and y-intercepts

To find the x-intercepts, we set \(f(x) = 0\) and solve for \(x\): \[0 = 3x^2 + 5x - 7\] This quadratic equation does not factor easily, so we can use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Plug in the values \(a = 3\), \(b = 5\), and \(c = -7\): \[x = \frac{-5 \pm \sqrt{5^2 - 4(3)(-7)}}{2(3)}\] \[x = \frac{-5 \pm \sqrt{109}}{6}\] The x-intercepts are \(\frac{-5 + \sqrt{109}}{6}\) and \(\frac{-5 - \sqrt{109}}{6}\). To find the y-intercept, we set \(x = 0\) and solve for \(f(x)\): \[f(x) = 3(0)^2 + 5(0) - 7\] \[f(x) = -7\] The y-intercept is at the point \((0, -7)\).

4. Plot key points and sketch the parabola

Now we have the vertex \(\left(-\frac{5}{6}, -\frac{85}{12}\right)\), the x-intercepts \(\frac{-5 + \sqrt{109}}{6}\) and \(\frac{-5 - \sqrt{109}}{6}\), and the y-intercept \((0, -7)\). Plot these key points on the coordinate plane and sketch the parabola. Since the parabola opens upwards, the graph will have a minimum value at the vertex and symmetric about the axis of symmetry \(x = -\frac{5}{6}\).

Key concepts

Parabola

A parabola is a U-shaped curve that can open upwards or downwards, and it is the graph of a quadratic function. This type of function takes the general form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our exercise, we encounter the function \( 3x^2 + 5x - 7 \), which reveals a parabola opening upwards because the coefficient \( a \) is positive. Understanding the shape and direction a parabola takes is fundamental to graphing quadratic functions.

When graphing, it's important to note key features such as the vertex, the axis of symmetry, and the intercepts. The vertex is essentially the highest or lowest point on the graph, the intercepts are where the parabola crosses the x-axis and y-axis, and the axis of symmetry is a vertical line that divides the parabola into two mirror images.

Vertex of a Parabola

The vertex of a parabola is the point where it turns; it is also the maximum or minimum point of the function. For a parabola \( f(x) = ax^2 + bx + c \) that opens upwards, like our exercise's quadratic function, the vertex is the lowest point on the graph. To find the vertex, we use a method called completing the square, which rewrites the function in vertex form \( f(x) = a(x-h)^2 + k \), where \( h \) and \( k \) are the x and y coordinates of the vertex, respectively.

Identifying the Vertex

For our exercise, after completing the square, we determine the vertex to be \( \left(-\frac{5}{6}, -\frac{85}{12}\right) \), which will be the starting point for plotting our graph.

X-intercepts

X-intercepts are the points where the parabola crosses the x-axis. These occur where \( f(x) = 0 \). To find the x-intercepts of a quadratic function, we can factor the function or use the quadratic formula when factoring is complex or not possible.

Finding the X-intercepts

In the given exercise, we apply the quadratic formula because the quadratic equation doesn't factor neatly. By substituting the values from our function into the quadratic formula, we get two x-intercepts: \( \frac{-5 + \sqrt{109}}{6} \) and \( \frac{-5 - \sqrt{109}}{6} \). These intercepts are essential for graphing as they provide specific points where the graph will touch the x-axis.

Y-intercept

The y-intercept is where the graph of the function intersects the y-axis. This happens when the input x is zero. To find the y-intercept, we simply set \( x = 0 \) in the function and solve for \( f(x) \).

Locating the Y-intercept

For the function \( 3x^2 + 5x - 7 \), setting \( x = 0 \) gives us the y-intercept at \( f(x) = -7 \), which corresponds to the point \( (0, -7) \) on the graph. It's a straightforward yet crucial point that helps anchor our parabola on the coordinate plane.

Axis of Symmetry

The axis of symmetry is a vertical line that runs through the vertex of the parabola and divides it into two symmetrical halves. For every parabola, this axis goes through the vertex's x-coordinate, \( h \). The equation of the axis of symmetry can be written as \( x = h \).

Determining the Axis of Symmetry

Given the vertex we previously found in our exercise, \( \left(-\frac{5}{6}, -\frac{85}{12}\right) \), the axis of symmetry is the line \( x = -\frac{5}{6} \). This helps us ensure that the parabola is symmetric when we sketch the graph.

Completing the Square

Completing the square is a technique used to rewrite a quadratic function in a form that makes it easier to identify key characteristics like the vertex and axis of symmetry. It involves creating a perfect square trinomial from the quadratic expression, which then simplifies to a squared binomial.

Applying Completing the Square

In our exercise, we start by factoring out the leading coefficient from the x terms and then add and subtract a constant to complete the square. This process reorganizes the initial equation into vertex form, allowing us to read off the vertex directly. It's a step-by-step process that allows us to convert the standard quadratic expression into a more manageable form.

Quadratic Formula

The quadratic formula offers a way to find the x-intercepts of a quadratic function when factoring is not feasible. It is derived from the general form of a quadratic equation \( ax^2 + bx + c = 0 \) and is stated as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The formula provides solutions for x that make the equation true, representing the x-intercepts of the function when graphed.

Utilizing the Quadratic Formula

In step 3 of our exercise, we used this formula by plugging in the coefficients from the function \( 3x^2 + 5x - 7 \) to find that the parabola crossed the x-axis at two points. Using the quadratic formula is a reliable method to ensure we capture all possible points where the parabola might intersect the x-axis.