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Algebra 1

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f(x)=x^2+3 Solve for f(2)and f(-5)

I need help, can someone please help me by explaining step by step how to solve this problem, I am most grateful to you. Thank you before hand.

  • Algebra 1 - ,

    just plug in the values for x and solve

    f(x) = x^2 + 3
    f(2) = 2^2 + 3
    f(2) = ?

    f(x) = x^2 + 3
    f(5) = 5^2 + 3
    f(5) = ?

  • Algebra 1 - ,

    Please forgive me, math is not my forte as you can see. Does that mean that f(5)= 5? and f(2)= 2? Thus, the answer is (5,2)or just 5 and 2? I guess what I'm trying to determing is how do you write the answer? Thank you and pleae excuse my confusion.

  • Algebra 1 - ,

    f(x) = x^2 + 3
    f(2) = 2^2 + 3
    f(2) = ?

    2^2 = 4
    f(2) = 2^2 + 3
    f(2) = 4 + 3
    f(2) = 7

    the answer for f(2) is 7
    when x = 2, the function value = 7

    now, you do the other one

  • Algebra 1 - ,

    Thank you for your quick response, and your much needed help. is the answer to the the second problem 10? Thank you.

  • Algebra 1 - ,

    No, f(5) does not = 10

    f(x) = x^2 + 3
    f(5) = 5^2 + 3
    f(5) = ?

    5^2 = ?
    5^2 + 3 = ?

    show me how you got 10

  • Algebra 1 - ,

    I added 5 and 2 + 3. I am obviously wrong. Sorry!

  • Algebra 1 - ,

    I think the answer is 5 because of the polynomial 2, I could be wrong but doesn't that amount to 5 twos, in other words, 2,2,2,2,2, which equal 5, one whole number? Sorry, i'm lost.

  • Algebra 1 - ,

    f(x) = x^2 + 3
    f(5) = 5^2 + 3
    f(5) = ?

    5^2 = 5 x 5 = 25
    f(5) = 5^2 + 3
    f(5) = 25 + 3
    f(5) = 28

  • Algebra 1 - ,

    5^2 is "5 squared."

    xxxxx
    xxxxx
    xxxxx
    xxxxx
    xxxxx
    (Not a perfect square there because of the text size).

    How many are in 5^2? (Just count them up. It is 5 up or 5 across in every row/column).



    )

  • Algebra 1 - ,

    Thank you. I wholeheartedly, appreciate your help. Your help does make a big difference in how to solve some of these algebra problems. Thank you, again.

  • Algebra 1 - ,

    you're very welcome :)

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