Chapter 3, Polynomial and Rational Functions - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 288: 8

a) Vertex: $(-1, -4)$ X-intercepts: $(\frac{2\sqrt 3 -3}{3},0)$ and $(\frac{-2\sqrt 3-3}{3}, 0)$ Y-intercepts: $(0,-1)$ b) $-4$ c) Domain: (- ∞, ∞) Range: [-4, ∞)

$f(x)=3x^2+6x-1$ a) The vertex of a quadratic function can be achieved by completing a square. $f(x)=3x^2+6x-1$ $f(x)=3(x^2+2x+1)-3-1$ $f(x)=3(x+1)^2-4$ In standard form $f(x)=a(x-h)^2+k$, the vertex's coordinates always take the form of $(h,k)$. Thus, the vertex's coordinates are $(-1, -4)$. Because $a>0$ $(3>0)$, $-4$ is the function's minimum value. Therefore, the graph does intersect the x-axis. At the x-axis, the y-coordinate is always zero. Hence, to get the coordinates of the x-intercepts, we need to solve for x when $f(x)=0$. $3(x+1)^2-4=0$ $3(x+1)^2=4$ $(x+1)^2=\frac{4}{3}$ $x+1=\frac{2\sqrt 3}{3}$ or $x+1=-\frac{2\sqrt 3}{3}$ $x=\frac{2\sqrt 3 -3}{3}$ or $x=\frac{-2\sqrt 3-3}{3}$ The x-intercepts are $(\frac{2\sqrt 3 -3}{3},0)$ and $(\frac{-2\sqrt 3-3}{3}, 0)$ All points on the y-axis have zero as the value of the x-coordinates. Replacing $x$ with zero on the equation would help us find the y-coordinate of the y-intercept. $y=f(0)=3*0^2+6*0-1=0+0-1=-1$ The y-intercept is $(0,-1)$ b) As stated above. c) The domain of all quadratic function is all real numbers. The range of a quadratic function depends on its minimum/maximum value. In this equation, because the minimum value is $-4$, the value of the function cannot get lower than that. Therefore, its range is [-4, ∞).