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# parabola

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The graph of y=ax^2+bx+c is given below, where a, b, and c are integers. Find a-b+c.

May 19, 2021

There is a falsity in the statement of this problem; a is definitely not an integer. It is even evident from the graph that a is around $$\frac{2}{3}$$, since if you start with the vertex and move one unit to the right the y value increases by less than a unit; b and c are, however integers, b = -4 and c =7. You can find these values by writing the vertex form $$y=a{(x-h)}^{2}+k$$ and plugging in 3 for h and 1 for k and expanding to get $$y = a({x}^{2}-6x+9) +1$$. Finally, since the point (0, 7) is on the graph, we have 7 = 9a +1 , which yields $$a =\frac{2}{3}$$. So $$b=\frac{2}{3}(-6)=-4$$ and $$c=\frac{2}{3}(9) + 1= 7$$. Finally, the parabola has equation
$$y=\frac{2}{3}{x}^{2}-4x+7.$$