dcowboys107 Posted September 11, 2009 Share Posted September 11, 2009 The graph of a quadratic function f(x) has its vertex at (5,-6) and passes through the point (2,-51). Find a formula for f(x) I got 15(x-5)^2 - 6 Why is this wrong? Find the minimum value of the function (5x-2)(2x+2). I got -49/20 after multiplying out the equation to get 5x^2+3x-2 (simplified). I did -3/10 and put it back into the equation and got it wrong. . . Any ideas? Link to comment Share on other sites More sharing options...
Bignose Posted September 11, 2009 Share Posted September 11, 2009 The graph of a quadratic function f(x) has its vertex at (5,-6) and passes through the point (2,-51). Find a formula for f(x) I got 15(x-5)^2 - 6 Why is this wrong? it is wrong because if you put 2 into the expression for x, it evaluates to 129, not -51. So, it clearly doesn't answer the question. Find the minimum value of the function (5x-2)(2x+2). I got -49/20 after multiplying out the equation to get 5x^2+3x-2 (simplified). I did -3/10 and put it back into the equation and got it wrong. . . Any ideas? What do you mean by "I did -3/10"? You mean you inserted the value of -3/10 for x? Why did you use that number? Also, (5x-2)(2x+2) isn't equal to 5x^2+3x-2. Also also, it isn't an "equation" because there isn't an equals sign. You have an "expression" or maybe a "function". Link to comment Share on other sites More sharing options...
dcowboys107 Posted September 12, 2009 Author Share Posted September 12, 2009 I got -3/10 by using -b/2a. I put it into the expression and got said number. How do you find the what the value is then? Link to comment Share on other sites More sharing options...
Bignose Posted September 12, 2009 Share Posted September 12, 2009 Maximums and minimums of f(x)=0 are found by taking the derivative of f(x) and setting it equal to zero and solving for the value(s) of x that solve f'(x)=0. If [math]f(x)=ax^2 + bx +c[/math] then [math]f'(x)=2ax+b[/math] which when you set equal to zero you will get that -b/2a. Your error stems from that fact that [math](5x-2)(2x+2) \ne 5x^2+3x-2[/math]. Link to comment Share on other sites More sharing options...
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