Consider the quadratic function [tex]f(x)=-x^2-2x+24[/tex]. Find exact answers for each of the following:

1. The [tex]x[/tex]-intercepts are [tex]\square[/tex]
2. The [tex]y[/tex]-intercept is [tex]\square[/tex]
3. The vertex is [tex]\square[/tex]
4. The line of symmetry has the equation [tex]\square[/tex]



Answer :

Let's consider the quadratic function [tex]\( f(x) = -x^2 - 2x + 24 \)[/tex].

1. Finding the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts are the points where the function [tex]\( f(x) \)[/tex] intersects the [tex]\( x \)[/tex]-axis. These occur where [tex]\( f(x) = 0 \)[/tex]. Thus, we need to solve the equation:
[tex]\[ -x^2 - 2x + 24 = 0 \][/tex]

Solving this quadratic equation, we find the [tex]\( x \)[/tex]-intercepts to be:
[tex]\[ x = -6 \quad \text{and} \quad x = 4 \][/tex]

Therefore, the [tex]\( x \)[/tex]-intercepts are [tex]\( (-6, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].

2. Finding the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept is the point where the function [tex]\( f(x) \)[/tex] intersects the [tex]\( y \)[/tex]-axis. This occurs when [tex]\( x = 0 \)[/tex]. So, we evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 - 2(0) + 24 = 24 \][/tex]

Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 24) \)[/tex].

3. Finding the vertex:
The vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the [tex]\( x \)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]

For the given function [tex]\( f(x) = -x^2 - 2x + 24 \)[/tex], the coefficients are [tex]\( a = -1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 24 \)[/tex]. Substituting these values into the formula:
[tex]\[ x = -\frac{-2}{2(-1)} = 1 \][/tex]

Now, we find the corresponding [tex]\( y \)[/tex]-coordinate by evaluating [tex]\( f(x) \)[/tex] at this [tex]\( x \)[/tex]-value:
[tex]\[ f(-1) = -(-1)^2 - 2(-1) + 24 = -1 + 2 + 24 = 25 \][/tex]

Therefore, the vertex is [tex]\( (-1, 25) \)[/tex].

4. Finding the line of symmetry:
The line of symmetry in a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the line [tex]\( x = -\frac{b}{2a} \)[/tex]. As we found earlier:
[tex]\[ x = -\frac{-2}{2(-1)} = 1 \][/tex]

Therefore, the line of symmetry of the given quadratic function is [tex]\( x = -1 \)[/tex].

Summarizing:

- The [tex]\( x \)[/tex]-intercepts are [tex]\( (-6, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 24) \)[/tex].
- The vertex is [tex]\( (-1, 25) \)[/tex].
- The line of symmetry is [tex]\( x = -1 \)[/tex].