? asked in Science & MathematicsMathematics · 1 decade ago

can anyone help me solve this quadratic equation? y=-x^2+1?

the only part im lost in is where is the linear term so i can find the axis and the vertex

4 Answers

Relevance
  • Anonymous
    1 decade ago
    Favorite Answer

    You know what x^2 looks like right? It's a parabola with the vertex on the origin. The negative sign inverts it so the parabola points down, the plus one moves it up 1 on the y-axis so the vertex is located at (0,1) and the parabola points downward.

    -omgyay_dotcom

    Source(s): College Alg. Trig, Cal I, Cal II, Cal III, Differential Equations...trust me :-P
  • 1 decade ago

    I assume you mean the axis of symmetry. Remember all quadratics are in the form y = ax^2 + bx + c. Since you have no bx term it means the b value must be 0.

    Really the equation should look like y = x^2 + 0x + 1

  • 1 decade ago

    Well, -x^2 + 1 has no middle term, so when its being solved, the middle one cancels out during its evaluation.

    This can be rewritten as y = -1x^2 + 0x + 1

    So, it would be simplified as y = (-x + 1)(x + 1)

    Expanding it, you get -x^2 + x - x + 1

    The x's in the middle cancels out, so you get y = -x^2 + 1

  • 1 decade ago

    well the basic rule of linear equations is y=mx+c so you got your gradient there, and your Y intercept but im wondering if this is actually what you want to find out, is it? If not tell me what you need

Still have questions? Get your answers by asking now.