jbaybeh
contestada

Use the parabola tool to graph the quadratic function f(x)=x2−12x+27.

Graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

Respuesta :

cerverusdante
Remember that for a quadratic function of the form [tex]a x^{2} +bx+c[/tex] we can find its vertex with formula: [tex]h= \frac{-b}{2a} [/tex] where [tex]h[/tex] is the x-coordinate of the vertex; then we will find the y-coordinate by replacing [tex]h[/tex] in our original function.

From the question know that [tex]a=1[/tex] and [tex]b=-12[/tex], so lets replace those values in our vertex formula to find the x-coordinate of our vertex:
[tex]h= \frac{-(-12)}{2(1)} [/tex]
[tex]h= \frac{12}{2} [/tex]
[tex]h=6[/tex]
Now lets replace that value into our original function to find the y-coordinate of our vertex:
[tex]f(6)=6^{2} -12(6)+27[/tex]
[tex]f(6)=36-72+27[/tex]
[tex]f(6)=-9[/tex]
Finally, we have our vertex (6,-9)

Now to graph our function we are going to take advantage of its line of symmetry; if the vertex is (6,-9) the line of symmetry of the parabola is x=6, so if we chose the point (6,0), our second point will have coordinates (3,0) as you can see in the picture.
Ver imagen cerverusdante
MrRoyal

The vertex of a parabola is the minimum or the maximum point of the parabola

The equation of the parabola is given as:

  • [tex]f(x) = x^2 + 12x + 27[/tex]

The x-coordinate of the vertex is calculated using:

[tex]x = -\frac{b}{2a}[/tex]

So, we have:

[tex]x = -\frac{12}{2 \times 1}[/tex]

[tex]x = -\frac{12}{2}[/tex]

[tex]x = -6[/tex]

Substitute -6 for x in f(x)

[tex]f(-6) = (-6)^2 + 12(-6) + 27[/tex]

[tex]f(-6) =-9[/tex]

So, the vertex of the parabola is (-6,-9)

Let x = 0.

So, we calculate f(0) as follows:

[tex]f(0) = (0)^2 + 12(0) + 27[/tex]

[tex]f(0) = 27[/tex]

This means that f(x) passes through (0,27)

See attachment of the parabola

Read more about parabolas at:

https://brainly.com/question/4148030

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