MATH SOLVE

5 months ago

Q:
# 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.

Accepted Solution

A:

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.

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.