Value: 2 $$ \begin{array}{l} ext { What is the vertex of the quadratic function given by the equation } \ f(x)=-(x+3)^{2}+5\end{array} $$

Question

Value: 2 $$ \begin{array}{l}	ext { What is the vertex of the quadratic function given by the equation } \ f(x)=-(x+3)^{2}+5\end{array} $$

UpStudy AI · Accepted Answer

The given quadratic function is in vertex form:

$$
f(x) = a(x - h)^2 + k
$$

where $$(h, k)$$ represents the vertex of the parabola.

Let's analyze your function:

$$
f(x) = -(x + 3)^2 + 5
$$

1. **Identify $$a$$, $$h$$, and $$k$$:**
   - $$a = -1$$ (which indicates the parabola opens downward)
   - The term inside the square is $$(x + 3)$$, which can be rewritten as $$(x - (-3))$$. So, $$h = -3$$.
   - $$k = 5$$

2. **Determine the vertex:**
   - The vertex $$(h, k)$$ is $$(-3, 5)$$.

**Therefore, the vertex of the quadratic function $$f(x) = -(x + 3)^2 + 5$$ is $$(-3, 5)$$.**
The vertex of the quadratic function $$ f(x) = -(x + 3)^2 + 5 $$ is $$(-3, 5)$$.

Value: 2

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Algebra Questions