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MariettaO [177]
1 year ago
13

- Polynomial Functions -For each function, state the vertex; whether the vertex is a maximum or minimum point; the equation of t

he axis of symmetry and whether the function's graph is steeper than, flatter than, or the same shape as the graph of f(x)=x²

Mathematics
1 answer:
vladimir2022 [97]1 year ago
3 0

EXPLANATION

Given the function f(x) = (x-6)^2 + 1

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

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

=x^2-12x+37\mathrm{The\: parabola\: params\: are\colon}a=1,\: b=-12,\: c=37x_v=-\frac{b}{2a}x_v=-\frac{\left(-12\right)}{2\cdot\:1}\mathrm{Simplify}x_v=6y_v=6^2-12\cdot\: 6+37

Simplify:

y_v=1

\mathrm{Therefore\: the\: parabola\: vertex\: is}\mleft(6,\: 1\mright)\mathrm{If}\: a\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}a=1\mathrm{Minimum}\mleft(6,\: 1\mright)\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

y=x^2-12x+37\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}a=1,\: b=-12x=\frac{-\left(-12\right)}{2\cdot\:1}\mathrm{Refine}

Axis of simmetry : x=6

The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.

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