Which of the following is the graph of f(x) = x2 − 6x + 5?

graph of a quadratic function with a minimum at 2, negative 9 and x intercepts at negative 1 and 5
graph of a quadratic function with a minimum at 3, negative 4 and x intercepts at 1 and 5
graph of a quadratic function with a minimum at 2.5, negative 2.4 and x intercepts at 1 and 4
graph of a quadratic function with a minimum at negative 1.5, negative 6.2 and x intercepts at 1 and negative 4

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.

Ver imagen SaniShahbaz

Answer:

The second one is the answer

Step-by-step explanation:

i got it correct on the test

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