Consider the graph of the quadratic function f(x)=2x^2-4x-7 Find: (a) the vertex, (b) the axis of symmetry. (c) the y-intercept, and (d) the x-intercepts, if any Round to two decimal places as needed. Type NA if none ex Provide your answer below: (a) the vertex= (Write as an ordered pair) (b) the axis of symmetry is the line x= (c) The y-intercept is found at y= (d) The x-intercept is found at x=

## Step 1: Find the vertex<br />### The vertex of a quadratic function $f(x) = ax^2 + bx + c$ is given by the formula $\left( \frac{-b}{2a}, f\left(\frac{-b}{2a}\right) \right)$. For $f(x) = 2x^2 - 4x - 7$, $a = 2$, $b = -4$, and $c = -7$. Calculate $\frac{-b}{2a}$ and then find $f\left(\frac{-b}{2a}\right)$.<br />\[<br />x = \frac{-(-4)}{2(2)} = \frac{4}{4} = 1<br />\]<br />\[<br />f(1) = 2(1)^2 - 4(1) - 7 = 2 - 4 - 7 = -9<br />\]<br />Thus, the vertex is $(1, -9)$.<br /><br />## Step 2: Find the axis of symmetry<br />### The axis of symmetry for a quadratic function $f(x) = ax^2 + bx + c$ is the vertical line $x = \frac{-b}{2a}$. From Step 1, we have $x = 1$.<br />\[<br />x = 1<br />\]<br /><br />## Step 3: Find the y-intercept<br />### The y-intercept is found by evaluating $f(x)$ at $x = 0$. For $f(x) = 2x^2 - 4x - 7$:<br />\[<br />f(0) = 2(0)^2 - 4(0) - 7 = -7<br />\]<br />Thus, the y-intercept is $y = -7$.<br /><br />## Step 4: Find the x-intercepts<br />### The x-intercepts are found by solving the equation $2x^2 - 4x - 7 = 0$. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:<br />\[<br />x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-7)}}{2(2)} = \frac{4 \pm \sqrt{16 + 56}}{4} = \frac{4 \pm \sqrt{72}}{4} = \frac{4 \pm 6\sqrt{2}}{4} = 1 \pm \frac{3\sqrt{2}}{2}<br />\]<br />Thus, the x-intercepts are approximately:<br />\[<br />x = 1 + \frac{3\sqrt{2}}{2} \approx 3.12 \quad \text{and} \quad x = 1 - \frac{3\sqrt{2}}{2} \approx -1.12<br />\]