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=-(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=-(b)/(2a)`
`x_v=-(\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<0,\: \mathrm{then\: the\: vertex\: is\: a\: maximum\: value}`
`\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}`
`a=1`
`\mathrm{Minimum}\mleft(6,\: 1\mright)`

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=(-b)/(2a)`
`a=1,\: b=-12`
`x=(-\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.