Question

graph the equation $y = x^2 - 6x + 5$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex. using the graph, determine the roots of the equation $x^2 - 6x + 5 = 0$. click to plot points. click points to delete them.

Explanation:

Step1: Find the roots by factoring

To find the roots of \(x^{2}-6x + 5=0\), we factor the quadratic. We need two numbers that multiply to \(5\) and add to \(- 6\). The numbers are \(-1\) and \(-5\). So, \(x^{2}-6x + 5=(x - 1)(x - 5)=0\).
Setting each factor equal to zero: \(x-1 = 0\) gives \(x = 1\) and \(x - 5=0\) gives \(x = 5\). So the roots are \(x = 1\) and \(x = 5\).

Step2: Find the vertex of the parabola \(y=x^{2}-6x + 5\)

The \(x\)-coordinate of the vertex of a parabola \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). For \(y=x^{2}-6x + 5\), \(a = 1\), \(b=-6\), so \(x=-\frac{-6}{2\times1}=3\).
To find the \(y\)-coordinate, substitute \(x = 3\) into the equation: \(y=(3)^{2}-6\times3 + 5=9-18 + 5=-4\). So the vertex is \((3,-4)\).

Step3: Find additional points

We can find two more points. Let's take \(x = 0\): \(y=0^{2}-6\times0 + 5 = 5\), so the point is \((0,5)\).
Let's take \(x = 2\): \(y=2^{2}-6\times2+5=4 - 12 + 5=-3\), so the point is \((2,-3)\).
Let's take \(x = 4\): \(y=4^{2}-6\times4 + 5=16-24 + 5=-3\), so the point is \((4,-3)\).
Now we can plot the points \((1,0)\), \((5,0)\), \((3,-4)\), \((0,5)\), \((2,-3)\) (or \((4,-3)\)) on the graph. The roots of the equation \(x^{2}-6x + 5 = 0\) are the \(x\)-intercepts of the graph of \(y=x^{2}-6x + 5\), which we found as \(x = 1\) and \(x = 5\).

Answer:

The roots of the equation \(x^{2}-6x + 5=0\) are \(x = 1\) and \(x = 5\).