Question

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

Answer 1

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=37$$x_v=-\frac{b}{2a}$$x_v=-\frac{\left(-12\right)}{2\cdot\:1}$$\mathrm{Simplify}$$x_v=6$$y_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=-12$$x=\frac{-\left(-12\right)}{2\cdot\:1}$$\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.