# Chapter 3, Polynomial and Rational Functions - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises: 7

(a) $f(x) = (x-1)^2+2$ (b) vertex: $(1, 2)$ x-intercepts: none y-intercept: $3$ (c) Refer to the image below for the graph. (d) Domain $(-\infty, +\infty)$ Range: $[2, +\infty)$

#### Work Step by Step

$\bf{\text{ (a) Standard Form}}$ Group the terms with the variable $x$ to obtain: $f(x) = (x^2-2x)+3$ Complete the square by adding $(-\frac{2}{2})^2=1$ inside the parentheses. Subtract $(-\frac{2}{2})^2=1$ outside the parentheses. $f(x) = (x^2-2x+1)+3-1$ Write the trinomial in factored form to obtain: $\color{blue}{f(x) = (x-1)^2 + 2}$ $\bf{\text{(b) Vertex, x and y-intercepts}}$ The quadratic function has $a=1$ so the parabola opens upward. The vertex is at $(1, 2)$, which is above the x-axis. This means that the function has no x-intercepts as the graph will never touch the x-axis. Solve for the y-intercept by solving for $f(0)$: $f(x) = (x-1)^2+2 \\f(0) = (0-1)^2+2 \\f(0) = (-1)^2+2 \\f(0) = 1+2 \\f(0)=3$ Thus, the y-intercept is $3$. $\bf{\text{(c) Graph}}$ The given function involves (i) a horizontal shift of 1 unit to the right and (ii) a vertical shift of 2 units upward of the parent function $y=x^2$. Refer to the attached image in the answer part above for the graph. $\bf{\text{(d) Domain and range}}$ The function is quadratic so its domain is the set of real numbers, $(-\infty, +\infty)$. The y-values are from $2$ and above so the range is $(2, +\infty)$.

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