Question

How do you find the x-intercepts, axis of symmetry, and vertex of y= -x^2 -4x +5 consider the equation in STANDARD form

Answer 1

To analyze the quadratic equation \( y = -x^2 - 4x + 5 \), we first want to rewrite it in standard form, which is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola.

### Step 1: Find the Vertex

We can find the vertex using the formula \( h = -\frac{b}{2a} \) where \( a \) and \( b \) are the coefficients from the standard form \( y = ax^2 + bx + c \):
- Here, \( a = -1 \) and \( b = -4 \).

Calculating \( h \):
\[
h = -\frac{-4}{2 \cdot -1} = \frac{4}{-2} = -2
\]

Next, we find \( k \) by substituting \( h \) back into the original equation:
\[
k = -(-2)^2 - 4(-2) + 5 = -4 + 8 + 5 = 9
\]

So, the vertex is \( (-2, 9) \).

### Step 2: Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex. It's given by the equation \( x = h \).
Thus, the axis of symmetry for this parabola is:
\[
x = -2
\]

### Step 3: Find the X-Intercepts

To find the x-intercepts, we set \( y = 0 \):
\[
0 = -x^2 - 4x + 5
\]
Rearranging gives us:
\[
x^2 + 4x - 5 = 0
\]

Now we can factor the quadratic:
\[
(x + 5)(x - 1) = 0
\]

Setting each factor to zero gives us the solutions:
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Thus, the x-intercepts are \( (-5, 0) \) and \( (1, 0) \).

### Summary

- **Vertex:** \( (-2, 9) \)
- **Axis of Symmetry:** \( x = -2 \)
- **X-Intercepts:** \( (-5, 0) \) and \( (1, 0) \)