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# 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

- Anonymous1 decade agoFavorite 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