Step-by-step explanation:
Considering the function
[tex]f\left(x\right)=-x^2+6x-5[/tex]
solving
[tex]-x^2+6x-5=0[/tex]
[tex]\mathrm{Quadratic\:Equation\:Formula:}[/tex]
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=-1,\:b=6,\:c=-5:\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}[/tex]
[tex]x=\frac{-6+\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}[/tex]
[tex]=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:5}}{-2\cdot \:1}[/tex]
[tex]=\frac{-6+\sqrt{16}}{-2\cdot \:1}[/tex]
[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}[/tex]
[tex]-6+\sqrt{16}=-\left(6-\sqrt{16}\right)[/tex]
[tex]=\frac{6-\sqrt{16}}{2}[/tex]
[tex]=\frac{6-4}{2}[/tex]
[tex]x=1[/tex]
Similarly,
[tex]x=\frac{-6-\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}[/tex]
[tex]=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:5}}{-2\cdot \:1}[/tex]
[tex]=\frac{-6-\sqrt{16}}{-2}[/tex]
[tex]=\frac{6+\sqrt{16}}{2}[/tex]
[tex]\:x=5[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=1,\:x=5[/tex]
Also check the attached graph below.
From the graph as shown below, it is clear that
[tex]\mathrm{Domain\:of\:}\:-x^2+6x-5\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}[/tex]
and
[tex]\mathrm{Range\:of\:}-x^2+6x-5:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:4\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:4]\end{bmatrix}[/tex]
Also
[tex]\mathrm{Axis\:interception\:points\:of}\:-x^2+6x-5:\quad \mathrm{X\:Intercepts}:\:\left(1,\:0\right),\:\left(5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-5\right)[/tex]
[tex]\mathrm{Vertex\:of}\:-x^2+6x-5:\quad \mathrm{Maximum}\space\left(3,\:4\right)[/tex]
Thus, from the figure, it is clear that a partial parabola opening down, passing through (1, 0) and (5, 0) with endpoints (0, -5) and (6, -5).
