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The function f(t)=t^2+12t-18 represents a parabolaPart A: rewrite the function in vertex formPart B: determine the vertex and indicate whether it is a maximum or a minimum on the graphPart C: determine the axis of symmetry for f(t)

Sagot :

Solution:

A general form o quadratic function is given by the following expression:

[tex]f(x)=ax^2+bx+c[/tex]

The vertex form of this quadratic function is given by the following expression:

[tex]f(x)=a(x-h)^2+k[/tex]

where (h,k) is the vertex of the quadratic function.

PART A:

To find the vertex form of the given function, we must complete the square of this function:

[tex]t^2+12t-18+6^2-6^2[/tex]

this is equivalent to:

[tex](t+6)^2-18-6^2[/tex]

that is:

[tex](t+6)^2-54[/tex]

so that, the vertex form of the given function is:

[tex]f(t)=(t+6)^2-54[/tex]

Part B: According to the previous part, we can conclude that the vertex is:

[tex](h,k)=(-6,-54)[/tex]

Part C: The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. So that, we can conclude that the axis of symmetry for f(t) is:

[tex]x\text{ =-6}[/tex]

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