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What is the axis of symmetry for y=2x^2+16x-5

Sagot :

Answer:

[tex]x=-4[/tex]

Step-by-step explanation:

Convert to vertex form by completing the square

[tex]y=2x^2+16x-5\\\\y=2(x^2+8x-\frac{5}{2})\\ \\y+2(\frac{37}{2})=2(x^2+8x-\frac{5}{2}+\frac{37}{2})\\ \\y+37=2(x^2+8x+16)\\\\y+37=2(x+4)^2\\\\y=2(x+4)^2-37[/tex]

Since the equation is now in the form of [tex]y=a(x-h)^2+k[/tex], the axis of symmetry is where [tex]x=h[/tex], so, the axis of symmetry is [tex]x=-4[/tex].

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