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If I know that the points (-1,6) and (2,3) are on the graph of the quadratic function f(x) = 2x^2 + bx + c, how do I determine b and c?

Thanks to anyone who helps.

Dan
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    $y=2x^2+bx+c$, if the two points are on this graph then both points must satisfy the eqn, get 1 eqn for each point, solve – Vikram Sep 13 '14 at 02:52

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Use the given points.. (-1,6) & (2,3)

You know that the function will become 6 when $x$ is (-1) and it will become 2 when $x$ is 3

In other words$$f(-1) = 6$$ $$f(2) = 3 $$

that data leads you to 2 different equations as

$$6 = 2(-1)^2 + (-b) + c $$ $$3 = 2(2)^2 + 2b + c $$

now solving the system of equation is all that remaining... its all up to u :)

isuru
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  • The part I'm getting stuck on is when I end up with: -5 = 2b + c. – Dan Sep 13 '14 at 03:02
  • u get 2 equations. $$ 4 = -b + c ---(1)$$ $$-5 = 2b + c ---(2)$$ $$ (2) - (1) ; (-5) - 4 = 2b + c - (-b) - c = 3b $$ $$ -9 = 3b $$
    now solve it for "b" . substitute it in either (1) or (2) & get C ..     – isuru Sep 13 '14 at 03:06
  • Could you explain the 2nd last line for me, I'm not quite understanding the how and why of what you did there. – Dan Sep 13 '14 at 03:17
  • Ok, and then what did you do to b and c? – Dan Sep 13 '14 at 03:23
  • I subtract the $1^{st}$ equation from the $2^{nd}$

    i denote it by the symbol $"(2) - (1)"$

    first subtract the the left side then right side and put the "=" sign in between.

    Its all about solving simultaneous equations

    i suggest you to search about it on the net if u dont get what im saying

    subtracting the left hand side ; $ (-5) - 4 = -9 $ subtracting the right hand side ; $ [2b + c] - [-b + c] = 3b $ and $$ -9 = 3b $$

     – isuru Sep 13 '14 at 03:25