To graph the quadratic function $f(x) = 2(x+5)^2 - 1$, follow these steps:

Step 1: Identify the vertex

The quadratic function is in vertex form, which is given by: $f(x) = a(x-h)^2 + k$ Here, the vertex is at $(h, k)$.

For $f(x) = 2(x+5)^2 - 1$, you can identify:

• $h = -5$
• $k = -1$

Thus, the vertex is $(-5, -1)$.

Step 2: Determine the direction of the parabola

The coefficient $a = 2$ is positive, so the parabola opens upwards. Since $a = 2$, the parabola is vertically stretched, meaning it is narrower than the standard parabola $f(x) = (x-h)^2 + k$.

Step 3: Find additional points

You can calculate a few more points by choosing $x$-values around the vertex:

• For $x = -4$: $f(-4) = 2(-4+5)^2 - 1 = 2(1)^2 - 1 = 1$ So, the point is $(-4, 1)$.

• For $x = -6$: $f(-6) = 2(-6+5)^2 - 1 = 2(-1)^2 - 1 = 1$ So, the point is $(-6, 1)$.

• For $x = 0$: $f(0) = 2(0+5)^2 - 1 = 2(5)^2 - 1 = 49$ So, the point is $(0, 49)$.

Step 4: Plot the graph

1. Plot the vertex $(-5, -1)$.
2. Plot the additional points such as $(-4, 1)$, $(-6, 1)$, and $(0, 49)$.
3. Draw the parabola, making sure it is symmetric around $x = -5$ and opening upwards.

Would you like me to generate a graph for this function or explain any step further?

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Here are 5 related questions:

1. How does changing the coefficient of $a$ affect the shape of the parabola?
2. What is the axis of symmetry for the given function?
3. Can you find the y-intercept for the function?
4. What happens to the graph if we change $k$ in the vertex form?
5. How would the graph look if $a$ was negative instead of positive?

Tip: The vertex form of a quadratic function is helpful because it directly provides the vertex, making it easier to graph.