Question - Analysis of a Quadratic Function Graph

To determine the coordinates of point \(Q\):

Point \(Q\) lies on the x-axis, which means the y-coordinate is 0.

Set the function \(f(x)=x^2+6x-5\) equal to 0 and solve for \(x\):

\(x^2+6x-5=0\)

\(x=\frac{-6\pm\sqrt{6^2-4(1)(-5)}}{2(1)}\)

\(x=\frac{-6\pm\sqrt{36+20}}{2}\)

\(x=\frac{-6\pm\sqrt{56}}{2}\)

\(x=\frac{-6\pm 2\sqrt{14}}{2}\)

\(x=-3\pm\sqrt{14}\)

Since \(Q\) is to the right of the y-axis, we take the positive value:

\(Q=(x, y)\)

\(Q=(-3+\sqrt{14}, 0)\)

To determine the maximum point \(P\):

The vertex of a parabola \(y=ax^2+bx+c\) is given by the formula \(x=-\frac{b}{2a}\).

In this function \(a=1\) and \(b=6\), so:

\(x_P=-\frac{6}{2(1)}\)

\(x_P=-3\)

Substitute \(x_P\) into the function to find \(y_P\):

\(y_P=(x_P)^2+6x_P-5\)

\(y_P=(-3)^2+6(-3)-5\)

\(y_P=9-18-5\)

\(y_P=-14\)

So the maximum point \(P\) is:

\(P=(x_P, y_P)\)

\(P=(-3, -14)\)