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anonymous

  • 5 years ago

y=x2+4 can someone help me wth this graph equation i need to find the vertex and the intercepts

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  1. anonymous
    • 5 years ago
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    do you mean y = x^2 + 4?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    so the standard parabola is y = mx+b where b is the y-intercept so in your equation y = x^2 + 4 so the y-intercept = 4, or (0,4)

  4. anonymous
    • 5 years ago
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    and the x-intercepts are (2,0) and (-2,0) by putting y=0

  5. dumbcow
    • 5 years ago
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    y=0^2 +4 y=4 y_intercept is 4 0 = x^2 +4 -4 = x^2 No real x_intercepts vertex: a=1, b=0, c=4 x=-b/2a x=0/2 = 0 y = 0^2+4 = 4 vertex: (0,4)

  6. anonymous
    • 5 years ago
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    oh pellet..looks lyk an off day

  7. anonymous
    • 5 years ago
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    dumbcows right

  8. anonymous
    • 5 years ago
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    thanks everyone

  9. anonymous
    • 5 years ago
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    i ate a dumbcow at mcdonalds once.... with cheese and a bun and extra pickles... it was mighty tasty :)

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