Chapter 5 Polynomials and Polynomial Functions - 5.1 Use Properties of Exponents - 5.1 Exercises - Mixed Review - Page 335: 60

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$y=3x(x+4)$ $y=3x^2+12x$ The function is in quadratic form $y=ax^2+bx+c$ with $a=3, b=12, k=0$ Find the x-coordinate: $x=-\frac{b}{2a}=-\frac{12}{2.3}=-2$ Substitute for x: $y=3(-2)^2+12(-2)=-12$ The vertex is $(-2,-12)$ Also the axis of symmetry is $x=-2$ Find some other points: $x=-4 \rightarrow y=0$ $x=0 \rightarrow y=0$ $x=1 \rightarrow y=15$ $x=-5\rightarrow y=15$