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[red and white kop!]

draw the graph of y=2x^2 + bx + 4 for different values of b. How does changing b affect the curve ax^2 +bx + c?
i drew several graphs and i cant really 'see' anything. ive compared coordinates of turning points etc. but i dont get it. any help?

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ 2x^{2}+bx+4, \ \implies \ y \ = \ \frac{-b \ \pm\sqrt{b^{2}-32}}{4}\)

\(\displaystyle b^{2}-32 \ \ge \ 0 \ \implies \ b^{2} \ \ge \ 32, \ |b| \ \ge \ 4\sqrt2, \ \implies \ b \ \ge \ 4\sqrt2, \ or \ b \ \le \ -4\sqrt2, \ for \ all \ reals.\)

\(\displaystyle Now, \ that \ should \ tell \ you \ something.\)

 

[soroban]

Hello, red and white kop!!

Draw the graph of \(\displaystyle y\:=\:2x^2 + bx + 4\) for different values of \(\displaystyle b.\)

How does changing \(\displaystyle b\) affect the curve \(\displaystyle y \:=\:ax^2 +bx + c\) ?

\(\displaystyle \text{We can write the equation in the form: }\:y \;=\;a\left(x + \tfrac{b}{2a}\right)^2 - \left(\tfrac{b^2-4ac}{4a}\right)\)

\(\displaystyle \text{The vertex is located at: }\:\left(-\tfrac{b}{2a},\;-(\tfrac{b^2-4ac}{4a}) \right)\)
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\(\displaystyle \text{As }b\text{ increases, the parabola is translated left and down.}\)
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\(\displaystyle \text{As }b\text{ decreases, the parabola is translated right and up.}\)