1. Let y = 3x2 + 4x - 5

Find the y-intercept, x-intercepts, the coordinates of the vertex, and sketch the graph.

Solve 3x2 + 4x - 5 > 0 for x.

To find the y-intercept, let x = 0

y = (0) 2 + 4(0) - 5

= -5

The constant term is always the y-intercept in the graph of any polynomial.

To find the roots or x-intercepts, let y = 0

0 = 3x2 + 4x - 5

and solve for x. To solve a quadratic, transpose all terms to one side, get a 0 on the other and hope that it factors. In this case, we already have a 0 on one side, but it doesn't factor. But it is set up for the quadratic formula

The radical simplifies

We can factor out a 2 from the two terms on top.

The factor of 2 on top goes into the 6 on the bottom 3 times.

The square root of 19 is approximately 4.36, so the two roots are approximately

and

The x coordinate of the vertex is

-b/2a = -4/2(3) = -4/6 = -2/3

The y coordinate is

f(-b/2a) = f(-2/3) = 3(-2/3)2 + 4(-2/3) - 5

= 3(4/9) - 4(2/3) - 5

= 4/3 - 8/3 - 5

= -4/3 - 5

= -6 and 1/3

If we draw the graph going through all these points, bottoming out at the vertex, we get

Solve 3x2 + 4x - 5 > 0 for x.

We divide the real number line into the intervals between the roots which were at approximately -2.12 and 0.79.

We pick a point in each interval. In the first interval, the simplest point to use would be -3

f(-3) = 3(-3)2 + 4(-3) - 5

= 3(9) - 4(3) - 5

= 27 - 12 - 5 = 10 > 0

so -3 is a solution. We darken in the numbers in the first inverval.

The easiest number to use in the next inverval is 0. We know that f(0) = -5, the y-intercept and -5 < 0, so 0 is not a solution, and so neither will any of the numbers in the second interval.

The simplest number to use in the last interval is 1

f(1) = 3(1)2 + 4(1) - 5

= 3 + 4 - 5

= 2 > 0

so 1 is a solution. So all of the numbers in the last interval will be solutions. We darken in the numbers in the first and last inverval, but do not darken in any of the numbers in the other intervals.

The endpoints are not solutions because at the endpoints we get answers that are equal to 0 which is not allowed in this case.