Solution: We see that<\/p>\n
y=14 whenever -2x^2 + 6x=0<\/p>\n
To find the two x-coordinates where y=14, we solve<\/p>\n
-2x^2+6x=0<\/p>\n
using the common factor<\/a> technique:<\/p>\n -2x(x-3)=0<\/p>\n which implies that<\/p>\n -2x=0 or x-3=0<\/p>\n and finally<\/p>\n x=0 or x=3<\/p>\n Therefore, y=14 when x=0 or when x=3.<\/p>\n Knowing the shape of a parabola (which y=-2x2<\/sup> + 6x + 14 is), we can draw a rough sketch. y=-2(1.5)^2+6(1.5)+14 Therefore, we have (1.5,18.5) as our vertex.<\/p>\n The A\/S, which means “axis of symmetry”, is the vertical line x=1.5, as shown in red. The graph’s maximum value is the y coordinate of the vertex: 18.5. The range must therefore be y\u226418.5, since the graph goes down from there. \u00a0The domain is all real numbers; it always is for quadratic equations, of which y=-2x2<\/sup>+6x+14 is one.<\/p>\n Partial factoring makes finding the vertex of a quadratic function very easy. I’m sure you’ll agree:)<\/p>\n Source: Nelson Foundations of Mathematics 11<\/em>.<\/p>\n
\n![](\"\/..\/partfact0.png\"\/)
\nNote the vertex is at the top because the lead coefficient (-2, in this case) is negative. The vertex will be midway between the two x values where y is 14: x=0 and x=3. Getting the average of those two values, we arrive at 1.5, which means that the vertex will be at (1.5, y). To find the y value of the vertex, we put 1.5 for x into the equation:<\/p>\n
\ny=18.5<\/p>\n