Write y=2x^2+8x+5 in vertex form. Please
1 answer:
--> y=2x^2+8x+5
Step 1: Group the first 2 terms together, separating them from the constant term.
--> (2 x X^2+{8}x X)+{5}
Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x2 must be 1.
--> 2 x (x^2+{4} x X) +{5}
Step 3: Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.
--> 2 x(X^2{4}x X +4-4)+ {5}
--> 2 x(X+{2}^2 x 4) + {5}
Step 4: Distribute and add constants.
--> 2 x(X+{2}^2 -8+5
--> 2 x(X+{2}^2 +{-3}
Now it is successfully in vertex form and can be easily graphed.
The vertex is at (-2,-3)
The parabola opens up and has a y-intercept at (0, 5)
Here is a graph of this parabola:
Step 1: Group the first 2 terms together, separating them from the constant term.
--> (2 x X^2+{8}x X)+{5}
Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x2 must be 1.
--> 2 x (x^2+{4} x X) +{5}
Step 3: Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.
--> 2 x(X^2{4}x X +4-4)+ {5}
--> 2 x(X+{2}^2 x 4) + {5}
Step 4: Distribute and add constants.
--> 2 x(X+{2}^2 -8+5
--> 2 x(X+{2}^2 +{-3}
Now it is successfully in vertex form and can be easily graphed.
The vertex is at (-2,-3)
The parabola opens up and has a y-intercept at (0, 5)
Here is a graph of this parabola:
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