Consider the Quadratic function $$ f(x)=x^{2}-5 x-6 $$. Its vertex is $$ (\square) $$ Its largest $$ x $$-intercept is $$ x=\square $$. Its $$ y $$-intercept is $$ y=\square $$

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We start with the quadratic function

$$
f(x)=x^2-5x-6.
$$

**1. Finding the vertex**

The vertex $$(h, k)$$ of a quadratic function $$ax^2+bx+c$$ is given by

$$
h = -\frac{b}{2a}.
$$

Here, $$a=1$$ and $$b=-5$$, so

$$
h = -\frac{-5}{2 \times 1} = \frac{5}{2}.
$$

Next, we find $$k$$ by evaluating $$f\left(\frac{5}{2}\right)$$:

$$
k = f\left(\frac{5}{2}\right) = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) - 6.
$$

Calculate $$\left(\frac{5}{2}\right)^2$$:

$$
\left(\frac{5}{2}\right)^2 = \frac{25}{4}.
$$

Now, compute the rest:

$$
-5\left(\frac{5}{2}\right) = -\frac{25}{2} = -\frac{50}{4},
$$
$$
- 6 = -\frac{24}{4}.
$$

Thus,

$$
k = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} = \frac{25 - 50 - 24}{4} = -\frac{49}{4}.
$$

So, the vertex is

$$
\left(\frac{5}{2}, -\frac{49}{4}\right).
$$

**2. Finding the $$ x $$-intercepts**

To find the $$ x $$-intercepts, set $$ f(x)=0 $$:

$$
x^2-5x-6=0.
$$

This quadratic factors as

$$
(x-6)(x+1)=0.
$$

Setting each factor equal to zero gives:

$$
x-6=0 \quad \text{or} \quad x+1=0,
$$

which implies

$$
x=6 \quad \text{or} \quad x=-1.
$$

The largest $$ x $$-intercept is

$$
x=6.
$$

**3. Finding the $$ y $$-intercept**

The $$ y $$-intercept is found by evaluating $$ f(0) $$:

$$
f(0)=0^2-5(0)-6=-6.
$$

Thus, the $$ y $$-intercept is

$$
y=-6.
$$

**Summary**

- The vertex is $$\left(\frac{5}{2}, -\frac{49}{4}\right)$$.
- The largest $$ x $$-intercept is $$ x=6 $$.
- The $$ y $$-intercept is $$ y=-6 $$.
The vertex is $$\left(\frac{5}{2}, -\frac{49}{4}\right)$$, the largest $$ x $$-intercept is $$ x=6 $$, and the $$ y $$-intercept is $$ y=-6 $$.

Consider the Quadratic function .
Its vertex is
Its largest -intercept is .
Its -intercept is

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Consider the Quadratic function . Its vertex is Its largest -intercept is . Its -intercept is