To analyze the quadratic function \( f(x) = x^2 + 5x - 6 \), we can first find the vertex and intercepts.
Step 1: Find the Vertex
The vertex of a parabola represented by \( f(x) = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
In this case, \( a = 1 \) and \( b = 5 \):
\[ x = -\frac{5}{2 \cdot 1} = -\frac{5}{2} = -2.5 \]
Now, we find the y-coordinate of the vertex by plugging \( x = -2.5 \) back into the function:
\[ f(-2.5) = (-2.5)^2 + 5(-2.5) - 6 \] \[ = 6.25 - 12.5 - 6 \] \[ = 6.25 - 18.5 = -12.25 \]
So, the vertex is \( \left(-2.5, -12.25\right) \).
Step 2: Find the Intercepts
Y-intercept: To find the y-intercept, set \( x = 0 \):
\[ f(0) = 0^2 + 5(0) - 6 = -6 \]
So, the y-intercept is \( (0, -6) \).
X-intercepts: To find the x-intercepts, set \( f(x) = 0 \):
\[ x^2 + 5x - 6 = 0 \]
We can factor this quadratic:
\[ (x + 6)(x - 1) = 0 \]
So, the x-intercepts are \( x + 6 = 0 \) (i.e., \( x = -6 \)) and \( x - 1 = 0 \) (i.e., \( x = 1 \)). Thus, the x-intercepts are \( (-6, 0) \) and \( (1, 0) \).
Step 3: Axis of Symmetry
The axis of symmetry of the parabola can be found using the x-coordinate of the vertex:
\[ x = -2.5 \]
The axis of symmetry is: \[ x = -\frac{5}{2} \]
Step 4: Domain and Range
Domain: The domain of a quadratic function is always all real numbers, so in interval notation:
\[ \text{Domain: } (-\infty, \infty) \]
Range: Since this parabola opens upwards (as \( a > 0 \)), the range starts from the y-coordinate of the vertex:
\[ \text{Range: } [-12.25, \infty) \]
Summary
- The axis of symmetry is: \( x = -\frac{5}{2} \)
- The domain is: \( (-\infty, \infty) \)
- The range is: \( [-12.25, \infty) \)
You can now use this information to sketch the graph.