Let's solve each part of the problem step by step:
1. Finding the Vertex of [tex]\( f(x) = x^2 + 10x - 9 \)[/tex]
The vertex of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( f(x) = x^2 + 10x - 9 \)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 10\)[/tex]
- [tex]\(c = -9\)[/tex]
Using the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{10}{2 \times 1} = -\frac{10}{2} = -5 \][/tex]
Now, substitute [tex]\( x = -5 \)[/tex] back into the function to find the y-coordinate (the value of [tex]\( f(x) \)[/tex] at [tex]\( x = -5 \)[/tex]):
[tex]\[ f(-5) = (-5)^2 + 10(-5) - 9 \][/tex]
[tex]\[ f(-5) = 25 - 50 - 9 \][/tex]
[tex]\[ f(-5) = -34 \][/tex]
So, the vertex of the function is [tex]\( (-5, -34) \)[/tex].
2. Finding the y-intercept of [tex]\( f(x) = x^2 + 10x - 9 \)[/tex]
The y-intercept is the value of the function when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 + 10 \times 0 - 9 \][/tex]
[tex]\[ f(0) = -9 \][/tex]
So, the y-intercept of the function is [tex]\( -9 \)[/tex].
3. Determining the direction in which the parabola opens
The direction of the parabola is determined by the coefficient [tex]\( a \)[/tex]:
- If [tex]\( a > 0 \)[/tex], the parabola opens upward.
- If [tex]\( a < 0 \)[/tex], the parabola opens downward.
In this case, [tex]\( a = 1 \)[/tex], which is greater than 0. Therefore, the parabola opens upward.
To summarize:
- The vertex of [tex]\( f(x) = x^2 + 10x - 9 \)[/tex] is [tex]\( (-5, -34) \)[/tex].
- The y-intercept of [tex]\( f(x) = x^2 + 10x - 9 \)[/tex] is [tex]\( -9 \)[/tex].
- The function [tex]\( f(x) = x^2 + 10x - 9 \)[/tex] opens upward.
