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Identify the vertex, axis of symmetry and min/max value of each16) f(x) = x² - 12x + 44

User PradeepK
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ANSWER

Vertex: (6, 8)

Axis of symmetry: x = 6

Minimum value: y = 8

Step-by-step explanation

The x-value of the vertex of a quadratic function with standard form:


f(x)=ax^2+bx+c

is:


x_v=(-b)/(2a)

In this function a = 1 and b = -12. The x-value of the vertex is:


x_v=-(-12)/(2)=-(-6)=6

To find the y-value of the vertex we have to replace x by xv in the function and solve:


\begin{gathered} y_v=x^2_v-12x_v+44 \\ y_v=6^2-12\cdot6+44 \\ y_v=36-72+44 \\ y_v=8 \end{gathered}

So the vertex is (6, 8)

The axis of symmetry is a vertical line that passes through the vertex, so it's x = 6.

This function has a minimum value, because a > 0 (positive) so the branches of the parabola go upwards. Therefore, the vertex is the minimum value of the function: y = 8

Identify the vertex, axis of symmetry and min/max value of each16) f(x) = x² - 12x-example-1
User Novawaly
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