Problem 35 - College Algebra

Question

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$f(x)-2 x^{2}+4 x-3$$

Step-by-Step Solution

Answer

The graph of the function \(f(x) = 2x^2 + 4x - 3\) is a parabola that opens upwards, and has the vertex (-1, -1), y-intercept -3 and roots are 0.5 and -3. The axis of symmetry is \(x = -1\). The domain is all real numbers \((-∞, +∞)\) and the range is \([-1, +∞)\).

Step 1: Simplify the Quadratic Function

The given function is in the standard form \(ax^2 + bx + c\), with \(a = 2\), \(b = 4\) and \(c = -3\). The model for the vertex form of a quadratic function is \(a(x - h)^2 + k\), where \((h, k)\) is the vertex of the function. We can calculate the vertex (h, k) with the formulas \(h = -\frac{b}{2a}\) and \(k = f(h)\). Calculating these values gives us \(h = -\frac{4}{2*2} = -1\) and plugging the value of h into the function, we get \(k = 2*(-1)^2 + 4*(-1) - 3 = -1\). So the function \(f(x)\) simplified and written in vertex form is: \(2(x - (-1))^2 - 1\). So, our vertex \((h, k)\) is \((-1, -1)\).

Step 2: Graph The Quadratic Function

Start by plotting the vertex point (h, k) = (-1,-1) on the graph. Then, since a > 0, the parabola opens upwards. To find more points, you can use the formula, if an original point is (x,y), a point (x+1, y+2a) belongs to the parabola. Then, plot those points and sketch a smooth curve to represent the quadratic function \(f(x) = 2x^2 + 4x - 3\) or \(f(x) = 2(x+1)^2 - 1\). Also plot the intercepts, when \(x=0, f(0)=-3\) which is the y-intercept and when \(f(x)=0\), \(x\) equals to the roots of the equation which are, \(x = 0.5\) and \(x = -3\)

Step 3: Determine the Axis of Symmetry

The axis of symmetry is defined by the equation \(x = h\), where \(h\) is the x-coordinate of the vertex. So for this quadratic function, the axis of symmetry is \(x = -1\).

Step 4: Determine the Domain and Range

The domain of a quadratic function is all real numbers, thus for this function \(f(x) = 2x^2 + 4x - 3\), the domain is \((-∞, +∞)\). The range of a quadratic function in the form \(f(x) = a(x -h)^2 + k\) that opens upwards is \([k, +∞)\). Because our k value is -1, the range for this function is \([-1, +∞)\).

Key Concepts

Vertex of a Parabola

Understanding the vertex is essential for graphing a quadratic function, as it marks the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards. To find the vertex of a parabola, we use the equations \(h = -\frac{b}{2a}\) and \(k = f(h)\), derived from the function's standard form \(ax^2 + bx + c\).

For the function \(f(x) = 2x^2 + 4x - 3\), calculating the vertex gives us \(h = -\frac{4}{2*2} = -1\) and \(k = 2*(-1)^2 + 4*(-1) - 3 = -1\). Therefore, the vertex of the parabola is \( (-1, -1) \). When you graph this function, begin by plotting the vertex, which serves as a foundational reference point to shape the curve of the parabola.

Axis of Symmetry

The axis of symmetry in a quadratic function is a vertical line that divides the parabola into two mirror-image halves. This axis passes directly through the vertex of the parabola. For any quadratic equation in the form \(ax^2 + bx + c\), the axis of symmetry formula is \(x = -\frac{b}{2a}\), where \(b\) and \(a\) are coefficients from the quadratic equation.

Referring to our example \(f(x) = 2x^2 + 4x - 3\), we have already computed the value \(h = -1\) when finding the vertex. Consequently, the axis of symmetry for this function is the line \(x = -1\). Plot this as a dashed line on the graph to aid in symmetrically placing points and to ensure accuracy when drawing the parabola.

Domain and Range of a Quadratic Function

The domain of a function refers to all possible input (x) values, whereas the range is the set of all possible output (y) values. For quadratic functions, the domain is always all real numbers, which we express as \( (-\text{∞}, +\text{∞}) \).

In our function \(f(x) = 2x^2 + 4x - 3\), as with all quadratic functions, the domain remains \( (-\text{∞}, +\text{∞}) \). However, the range is dependent on whether the parabola opens upwards or downwards. For our function that opens upwards (since the leading coefficient \(a = 2\) is positive), the range begins at the y-coordinate of the vertex \(k\) and extends to positive infinity. Hence, the range for this quadratic function is \( [-1, +\text{∞}) \). This means that the output values start at -1 and increase without bound.