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Is the only way to solve this question to substitute points or THERES some other way?

10 Answers

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First, the coefficient of x^2 is negative, which means the parabola opens downwards, eliminating D. Second, the constant term is the y-intercept. The only one left that passes through (0, 1) is C.
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When you have a parabola with the equation

    y = Ax² + Bx + C

Then each of the three parameters have a specific meaning

    C = y-coordinate of intersection with the y-axis
    B = slope of curve when it intersects the y-axis (rarely mentioned)

    A = height difference between vertex and the points Δx=1 next to it
        positive if the parabola opens upwards, negative if it opens downwards

With these in mind, which parabola fits?
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You could absolutely use substitution, but I think it would be a waste of your time. A better question is, what each variable in the equation stands for.

Ax^2 + Box + C

The first one I look for is C; the y-intercept.

B is the slope at the y intercept

Based on this, you can safely say that C is the right answer
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You can also rewrite the function (now) general form into the   vertex form y = a(x - p)² + q and then the vertex should be (p,q)
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Well substitution is by far the easiest (for example x=0 and x=-1 narrows it down to c un this case, those are not hard to compute, or if you happen to know that the direction of the parabola depends on the sign of the coefficient of x^2 than just x=0 is enough).

There are a lot of other methods though (many of which have been mentioned already):

You can look at where the function intersects the axes

You can look at it's derivative, find its slope at certain points

You can look for its roots with the quadratic formula

You can look for the roots of its derivative and find its maximum/minimum values
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In the class on this, what solution method(s) did they teach?  Use one of those. (I’m not being snarky, just saying the instructor expects to see it answered the way taught.)
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x^2 = positive smile of graph

-x^2 = negative/sad mouth of graph

+1 is where f(x) cuts through y-axis
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are u from Kurdistan ?
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Or possibly through the use of a graphing calculator you could put the equation into "y=" and then it GRAPH and match the graph on the screen to the one on paper.
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When x=0 f(x) = 1.  Graph A fits.

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