Question

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

Answer 1

Answer:

• The x-intercepts will be:

        $\left(5,\:0\right),\:\left(1,\:0\right)$

• $\mathrm{Minimum}\space\left(3,\:-4\right)$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^2\:-\:6x\:+\:5$

Determining x-intercepts:

x-intercept is a point on the graph where y = 0.

$\mathrm{Solve\:}\:x^2-6x+5=0$

$x^2-6x+5=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=5:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}$

$x=\frac{-\left(-6\right)+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}$

  $=\frac{6+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}$

  $=\frac{6+\sqrt{16}}{2\cdot \:1}$          ∵ $6+\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}=6+\sqrt{16}$

  $=\frac{6+4}{2}$

  $=\frac{10}{2}$

  $=5$

and

$x=\frac{-\left(-6\right)-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}$

 $=\frac{6-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}$

 $=\frac{6-\sqrt{16}}{2\cdot \:1}$          ∵ $6-\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}=6-\sqrt{16}$

 $=\frac{6-4}{2}$

 $=\frac{2}{2}$

 $=1$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=5,\:x=1$

Therefore, the x-intercepts will be:

$\left(5,\:0\right),\:\left(1,\:0\right)$

Determining Minimum:

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{The\:parabola\:params\:are:}$

$a=1,\:b=-6,\:c=5$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

$\mathrm{Plug\:in}\:\:x_v=3\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=3^2-6\cdot \:3+5$

$y_v=-4$

$\mathrm{Therefore\:the\:parabola\:vertex\:is}$

$\left(3,\:-4\right)$

as

• $\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}$

• $\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

but

$a=1$

Therefore,

• $\mathrm{Minimum}\space\left(3,\:-4\right)$

The graph is also attached below.

Answer 2

Answer:

The second one is the answer

Step-by-step explanation:

i got it correct on the test