Given the function in standard form, determine all the indicated information WITHOUT USING YOUR CALCULATOR OR DESMOS.COM. \[ f(x)=2 x^{2}+4 x-5 \] 1. Does the parabola open up or down? Concave up Concave down 2. Is the vertex a maximum or a minimum? Maximum Minimum 3. What is the $y$-intercept? Write answer as an $(x, y)$ ordered pair. 4. What is the $x$ coordinate of the vertex? Hint: use formula $x=\frac{-b}{2 a}$ \[ x= \] 5. What is the $y$ coordinate of the vertex? Hint: Substitute (plug in) the $x$ coordinate you found for the vertex into the original equation. \[ y= \]

Step 1 ：The given function is a quadratic function in the form of \(f(x) = ax^2 + bx + c\).

Step 2 ：The coefficient of \(x^2\) is positive, so the parabola opens upwards.

Step 3 ：The vertex of a parabola that opens upwards is a minimum.

Step 4 ：The y-intercept of the function is the value of the function at \(x=0\). So, the y-intercept is \(-5\).

Step 5 ：The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\). So, the x-coordinate of the vertex is \(-1\).

Step 6 ：The y-coordinate of the vertex can be found by substituting the x-coordinate of the vertex into the function. So, the y-coordinate of the vertex is \(-7\).

Step 7 ：Final Answer: The parabola opens \(\boxed{up}\).

Step 8 ：Final Answer: The vertex is a \(\boxed{minimum}\).

Step 9 ：Final Answer: The y-intercept is \(\boxed{(-5)}\).

Step 10 ：Final Answer: The x-coordinate of the vertex is \(\boxed{-1}\).

Step 11 ：Final Answer: The y-coordinate of the vertex is \(\boxed{-7}\).