SOLUTION: y=2x^2+4X-6, find x&y intercepts, coordinates of vertex,graph equation

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Question 141449: y=2x^2+4X-6, find x&y intercepts, coordinates of vertex,graph equation
Answer by nabla(475) About Me  (Show Source):
You can put this solution on YOUR website!
To find the intercepts, we will first set x=0 and then y=0:
y=-6 <--- y-intercept
0=2x%5E2%2B4x-6
Now we need to solve the second equation:
Factor:
0=%282x-2%29%28x%2B3%29
This can only be when x=1 or x=-3. <---x-intercepts

The vertex is where the function is at a minimum. I will give you a little calculus after this, but it suffices to say that the vertex is x=-b/(2a) when given a polynomial of form ax^2+bx+c. So the vertex per that formula is: x=-4/4=-1
which gives a y value of 2-4-6=-8. The ordered pair, then, is (-1,-8).
Calculus:
We find the critical points:
f'(x)=4x+4
0=4x+4
-->x=-1
We then proceed as before, finding that the ordered pair is (-1,-8).

Graph:
graph%28+300%2C+200%2C+-5%2C+5%2C+-10%2C+10%2C+2x%5E2%2B4x-6+%29