+253 A quadratic function is given. Find the vertex and x-intercept.
+129852 Note sally1, that we can use something called the "discriminant" to see if this actually has any "real" roots.
If ..... b^2 - 4ac < 0 ....we have no real roots......and thus no "x intercepts"
Therefore
b^2 - 4ac =
4^2 - 4(2)(3) =
16 - 24 =
-8 .....and this is < 0 .....so there are no x intercepts
This is always a good thing to test first...it may save a lot of work.....!!!
+118677 Think about the quadratic formula Sally.
$$\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\
\mbox{this could be also written as}\\\\
x=\frac{-b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\\\\$$
These values of x will give you the two roots. And remember that parabolas are symmetrical.
if you think about this for a moment. maybe put the 2 points onto a number line.
$$\\\mbox{You will see that the axis of symmetry must be } x=\frac{-b}{2a}$$
so lets look at your problem
f(x) = 2x 2 + 4x + 3
$$\\\mbox{axis of symmetry is }x=\frac{-b}{2a}=\frac{-4}{4} =-1\\\\$$
$$\\When\:\: x=-1\qquad y=2(-1)^2+4*-1+3=2-4+3=1\\\\
\mbox{So the vertex is (-1,1)}$$
The roots are found by putting a=2, b=4 and c=3 into the quadratic formula.
You can probably do this yourself Sally.
NOTE: If you'd wanted to you could have found the roots first and then taken the average of them to get the axis of symmetry.
See if you can finish it yourself. Any questions just ask again. 
+129852 Note sally1, that we can use something called the "discriminant" to see if this actually has any "real" roots.
If ..... b^2 - 4ac < 0 ....we have no real roots......and thus no "x intercepts"
Therefore
b^2 - 4ac =
4^2 - 4(2)(3) =
16 - 24 =
-8 .....and this is < 0 .....so there are no x intercepts
This is always a good thing to test first...it may save a lot of work.....!!!
