Can you convert a quadratic function to vertex form with the quadratic formula or do you have to complete the square?
How do you do this?
For example, can you convert y = x^2 - 5x - 9 to vertex form with the quadratic formula?
Please show your work. 10 points to the best answer.
How do you do this?
For example, can you convert y = x^2 - 5x - 9 to vertex form with the quadratic formula?
Please show your work. 10 points to the best answer.
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You can't use the quadratic formula to obtain the vertex form.
However, you can derive how to get the vertex form from the quadratic equation.
Take your general quadratic equation.
y = ax^2 + bx + c
Complete the square.
y = a(x^2 + [b/a]x) + c
y = a(x^2 + [b/a]x + [b^2/(4a^2)]) + c - b^2/(4a)
y = a(x + b/(2a))^2 + (4ac - b^2)/(4a)
The vertex form is
y = a(x - h)^2 + k
Which means you can obtain h and k easily.
h = -b/(2a)
k = (4ac - b^2)/(4a)
However, you can derive how to get the vertex form from the quadratic equation.
Take your general quadratic equation.
y = ax^2 + bx + c
Complete the square.
y = a(x^2 + [b/a]x) + c
y = a(x^2 + [b/a]x + [b^2/(4a^2)]) + c - b^2/(4a)
y = a(x + b/(2a))^2 + (4ac - b^2)/(4a)
The vertex form is
y = a(x - h)^2 + k
Which means you can obtain h and k easily.
h = -b/(2a)
k = (4ac - b^2)/(4a)
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y = x^2 - 5x - 9
= x^2 -5x + (-5/2)^2 -9 - (-5/2)^2
= (x -5/2)^2 -9 - 25/4
= (x -5/2)^2 - 36/4 - 25/4
= ( x -5/2)^2 - 61/4
= x^2 -5x + (-5/2)^2 -9 - (-5/2)^2
= (x -5/2)^2 -9 - 25/4
= (x -5/2)^2 - 36/4 - 25/4
= ( x -5/2)^2 - 61/4