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