y=x2-5x+1
The equation you've provided is a quadratic function in the standard form:
\[ y = ax^2 + bx + c \]
Here, \( a = 1 \), \( b = -5 \), and \( c = 1 \).
To analyze this quadratic function, you might be interested in several key features:
Vertex: The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{-5}{2 \cdot 1} = \frac{5}{2} = 2.5 \]
To find the corresponding \( y \)-value (the value of the function at \( x = 2.5 \)):
\[ y = (2.5)^2 - 5(2.5) + 1 \] \[ y = 6.25 - 12.5 + 1 = -5.25 \]
So, the vertex of the parabola is \( (2.5, -5.25) \).
Axis of symmetry: The axis of symmetry for a parabola in standard form is the vertical line that passes through the vertex, which is \( x = 2.5 \).
Y-intercept: The y-intercept occurs when \( x = 0 \):
\[ y = (0)^2 - 5(0) + 1 = 1 \]
So, the y-intercept is \( (0, 1) \).
X-intercepts (roots): To find the x-intercepts, set \( y = 0 \):
\[ 0 = x^2 - 5x + 1 \]
You can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \] \[ = \frac{5 \pm \sqrt{25 - 4}}{2} \] \[ = \frac{5 \pm \sqrt{21}}{2} \]
So the x-intercepts can be calculated as:
\[ x_1 = \frac{5 + \sqrt{21}}{2} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{21}}{2} \]
These characteristics outline the basic features of the quadratic function \( y = x^2 - 5x + 1 \). If you have any specific aspects you'd like to explore further, let me know!
