Let's break down the problem step by step for the quadratic equation $y = x^2 - 4x - 5$.

1. Axis of Symmetry

For a quadratic equation of the form $y = ax^2 + bx + c$, the axis of symmetry is given by the formula:

$x = \frac{-b}{2a}$

Here, $a = 1$, $b = -4$, and $c = -5$. Substituting into the formula:

$x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2$

So, the axis of symmetry is:

$x = 2$

2. Vertex

The vertex is the point where the parabola changes direction. The x-coordinate of the vertex is the same as the axis of symmetry, which is $x = 2$. To find the y-coordinate of the vertex, substitute $x = 2$ into the original equation $y = x^2 - 4x - 5$:

$y = (2)^2 - 4(2) - 5 = 4 - 8 - 5 = -9$

So, the vertex is:

$(x, y) = (2, -9)$

3. Y-Intercept

The y-intercept occurs when $x = 0$. Substitute $x = 0$ into the equation:

$y = (0)^2 - 4(0) - 5 = -5$

So, the y-intercept is:

$(x, y) = (0, -5)$

4. X-Intercepts

The x-intercepts occur when $y = 0$. To find the x-intercepts, we solve $x^2 - 4x - 5 = 0$. Factor the quadratic equation:

$(x - 5)(x + 1) = 0$

Setting each factor to 0:

$x - 5 = 0 \quad \Rightarrow \quad x = 5$ $x + 1 = 0 \quad \Rightarrow \quad x = -1$

So, the x-intercepts are:

$(5, 0) \quad \text{(larger x-value)}$ $(-1, 0) \quad \text{(smaller x-value)}$

5. Graph of the Quadratic Equation

To graph the equation $y = x^2 - 4x - 5$, we use the vertex, axis of symmetry, and intercepts found above. The parabola will open upwards because the coefficient of $x^2$ is positive.

• The vertex is at $(2, -9)$.
• The axis of symmetry is the vertical line $x = 2$.
• The y-intercept is $(0, -5)$.
• The x-intercepts are $(-1, 0)$ and $(5, 0)$.

Would you like me to graph this for you as well?

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5 Related Questions:

1. How do you derive the formula for the axis of symmetry?
2. What effect does changing the coefficient of $x^2$ have on the graph?
3. Can we graph quadratic equations that don’t factor easily?
4. What happens to the vertex if you change the constant in a quadratic equation?
5. How do you use completing the square to find the vertex?

Tip:

The vertex form of a quadratic equation, $y = a(x-h)^2 + k$, makes it easy to identify the vertex $(h, k)$. Converting to this form helps in graphing.