Questions: Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the parabola's axis of symmetry. Use the graph to determine the function's domain and range. f(x)=2x^2+4x-1

Solution Steps

The vertex form of a quadratic function is given by: \[ f(x) = a(x-h)^2 + k \] where \((h, k)\) is the vertex. To find the vertex of the given quadratic function \(f(x) = 2x^2 + 4x - 1\), we use the formula for the x-coordinate of the vertex: \[ h = -\frac{b}{2a} \] where \(a = 2\) and \(b = 4\).

\[ h = -\frac{4}{2 \times 2} = -\frac{4}{4} = -1 \]

Substitute \(x = -1\) back into the function to find \(k\): \[ f(-1) = 2(-1)^2 + 4(-1) - 1 = 2(1) - 4 - 1 = 2 - 4 - 1 = -3 \]

Thus, the vertex is \((-1, -3)\).

Y-intercept:

The y-intercept occurs when \(x = 0\): \[ f(0) = 2(0)^2 + 4(0) - 1 = -1 \] So, the y-intercept is \((0, -1)\).

X-intercepts:

To find the x-intercepts, set \(f(x) = 0\): \[ 2x^2 + 4x - 1 = 0 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 4\), and \(c = -1\): \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \times 2 \times (-1)}}{2 \times 2} \] \[ x = \frac{-4 \pm \sqrt{16 + 8}}{4} \] \[ x = \frac{-4 \pm \sqrt{24}}{4} \] \[ x = \frac{-4 \pm 2\sqrt{6}}{4} \] \[ x = \frac{-2 \pm \sqrt{6}}{2} \]

Thus, the x-intercepts are \(\left(-1 + \frac{\sqrt{6}}{2}, 0\right)\) and \(\left(-1 - \frac{\sqrt{6}}{2}, 0\right)\).

The axis of symmetry for a parabola in the form \(f(x) = ax^2 + bx + c\) is the vertical line that passes through the vertex. Therefore, the axis of symmetry is: \[ x = -1 \]

Domain:

The domain of any quadratic function is all real numbers: \[ (-\infty, \infty) \]

Range:

Since the parabola opens upwards (as \(a = 2 > 0\)), the range is all real numbers greater than or equal to the y-coordinate of the vertex: \[ [-3, \infty) \]

Final Answer