We wish to construct the graph of y = 2 t^2 + 4 t - 3.

• We know that the graph is a parabola but we don't have any idea where on the x-y plane the graph lies.
• Until we know where the vertex and axis of symmetry are, it's just about pointless (not to mention time-consuming) to just start mindlessly plugging in numbers to create a table.

So we proceed by using what the quadratic formula tells us.

We know that a quadratic function has a parabolic graph.

• The central point of a parabola is its vertex.
• If we can locate the vertex of the parabola, it gives us a good idea where to locate the graph.

Since we don't yet know how to locate vertex, the next-best thing is to locate the zeros, the points where the graph crosses the t axis.

• On the t axis we clearly have y = 0, so the zeros occur wherever y = 0.
• If y = 0, then 2 t^2 + 4t - 3 = 0, so we find zeros by solving this equation.

As seen on the right side of the figure below, the quadratic formula gives us two solutions to the equation 2 t^2 + 4 t - 3 = 0.

These solutions tell us that zeros are at

t = (-4 +- `sqrt(40) ) / 4

= (-4 +- 6.3) / 4 (approximately)

= -1 +- 1.6 (approx.)

We note that the exact solutions, involving radicals, are - 1 +- `sqrt(10) / 2. However when dealing with real-world models we are almost always dealing with approximations, so we're going to wait until later to worry about expressing exact solutions.

We plot these zeros on the t axis:

• We first locate the point t = -1, then we move 1.6 units to the right and 1.6 units to the left ot t = -1

• Note that in the figure below, 1.58 is used instead of 1.6.
• This use of 1.58 is in error. 
• Having rounded the square root of 40 to 6.3, with two significant figures, it makes no sense to use three significant figures in the final result.  We should have used the two-significant-figure number 1.6.

We see that the parabola will therefore behave as follows.  It will

• pass through the zero points we have plotted
• descend with a more and more gradual slope to a vertex between these points
• reach a slope of zero at the vertex, then
• rise more and more steeply on the other side of the zeros.

We know that a parabola is symmetric about its vertex.

• This implies that the vertex must lie along an axis which is halfway between the zeros.
• The reason we wrote the two solutions to the quadratic equation as t = -1 +- 1.6 was to emphasize that t = -1 lies halfway between the zeros.
• The axis of symmetry of the parabola therefore lies at t = -1.