They-value at the absolute minimum point is
ymin= 2 xmin^2 + 4 xmin+ 3
= 2 × (−1)^2 + 4 × (−1)+ 3
= 2 − 4 + 3
=− 2 + 3
= 1
From this, we know that the coordinates of the vertex are (−1,1). As the basis for our next point, let’s
choosex=−3. That’s 2 units smaller than the x-value of the absolute minimum. We can plug that into
the function to get
y= 2 x^2 + 4 x+ 3
= 2 × (−3)^2 + 4 × (−3)+ 3
= 18 − 12 + 3
= 6 + 3
= 9
This gives us (−3, 9) as the coordinates of a second point on the curve. Finally, let’s pick an x-value that’s
2 units larger than xmin; that would be x= 1. Plugging it in, we obtain
Figure 24-10 Approximate graph of y= 2 x^2 + 4 x+ 3.
On both axes, each increment represents
1 unit.xy(–3,9) (1,9)Each axis
increment
is 1 unit(–1,1)No Real Zeros 411