1. **Stating the problem:** We are given the quadratic function $f(x) = 2(x+1)^2 + 1$ and asked to find the vertex coordinates. 2. **Formula and rules:** The vertex form of a quadratic function is $f(x) = a(x-h)^2 + k$, where $(h,k)$ is the vertex. 3. **Identify vertex:** Here, $a=2$, $h=-1$, and $k=1$ directly from the formula. 4. **Answer:** The vertex of the parabola is at the point $\boxed{(-1, 1)}$. --- 1. **Stating the problem:** Convert $f(x) = 2(x+1)^2 + 1$ to the general form $ax^2 + bx + c$. 2. **Formula:** Expand the square: $(x+1)^2 = x^2 + 2x + 1$. 3. **Intermediate work:** $$f(x) = 2(x^2 + 2x + 1) + 1 = 2x^2 + 4x + 2 + 1 = 2x^2 + 4x + 3$$ 4. **Answer:** The general form is $\boxed{2x^2 + 4x + 3}$. --- 1. **Stating the problem:** Find the intersection of $f(x) = 2x^2 + 4x + 3$ with the $y$-axis. 2. **Rule:** The $y$-axis corresponds to $x=0$. 3. **Calculation:** $$f(0) = 2(0)^2 + 4(0) + 3 = 3$$ 4. **Answer:** The graph intersects the $y$-axis at $\boxed{(0, 3)}$. --- 1. **Stating the problem:** Find intervals where $f(x) = 2(x+1)^2 + 1$ is increasing or decreasing. 2. **Rule:** Since $a=2 > 0$, parabola opens upwards. Vertex at $x=-1$ is minimum point. 3. **Intervals:** - Decreasing on $(-\infty, -1)$ - Increasing on $(-1, +\infty)$ 4. **Answer:** $f$ is decreasing on $(-\infty, -1)$ and increasing on $(-1, +\infty)$. --- "slug":"quadratic vertex", "subject":"algebra", "svg":"", "desmos":{"latex":"y=2(x+1)^2+1","features":{"intercepts":true,"extrema":true}}, "q_count":1