Respuesta :
Answer:
- The x-intercepts will be:
[tex]\left(5,\:0\right),\:\left(1,\:0\right)[/tex]
- [tex]\mathrm{Minimum}\space\left(3,\:-4\right)[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)\:=\:x^2\:-\:6x\:+\:5[/tex]
Determining x-intercepts:
x-intercept is a point on the graph where y = 0.
[tex]\mathrm{Solve\:}\:x^2-6x+5=0[/tex]
[tex]x^2-6x+5=0[/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{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]
[tex]x=\frac{-\left(-6\right)+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]
[tex]=\frac{6+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]
[tex]=\frac{6+\sqrt{16}}{2\cdot \:1}[/tex] ∵ [tex]6+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}=6+\sqrt{16}[/tex]
[tex]=\frac{6+4}{2}[/tex]
[tex]=\frac{10}{2}[/tex]
[tex]=5[/tex]
and
[tex]x=\frac{-\left(-6\right)-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]
[tex]=\frac{6-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]
[tex]=\frac{6-\sqrt{16}}{2\cdot \:1}[/tex] ∵ [tex]6-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}=6-\sqrt{16}[/tex]
[tex]=\frac{6-4}{2}[/tex]
[tex]=\frac{2}{2}[/tex]
[tex]=1[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=5,\:x=1[/tex]
Therefore, the x-intercepts will be:
[tex]\left(5,\:0\right),\:\left(1,\:0\right)[/tex]
Determining Minimum:
[tex]\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}[/tex]
[tex]\mathrm{The\:parabola\:params\:are:}[/tex]
[tex]a=1,\:b=-6,\:c=5[/tex]
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{\left(-6\right)}{2\cdot \:1}[/tex]
[tex]x_v=3[/tex]
[tex]\mathrm{Plug\:in}\:\:x_v=3\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}[/tex]
[tex]y_v=3^2-6\cdot \:3+5[/tex]
[tex]y_v=-4[/tex]
[tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}[/tex]
[tex]\left(3,\:-4\right)[/tex]
as
- [tex]\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex]
- [tex]\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex]
but
[tex]a=1[/tex]
Therefore,
- [tex]\mathrm{Minimum}\space\left(3,\:-4\right)[/tex]
The graph is also attached below.

Answer:
The second one is the answer
Step-by-step explanation:
i got it correct on the test