A quadratic polynomial y = f(x) with absolute | vedclass

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(D) Given $f(x) = ax^2 + bx + c$. Since the absolute term is $3$,$c = 3$. Since the leading coefficient is unity,$a = 1$. Thus,$f(x) = x^2 + bx + 3$.S

Vedclass · Accepted Answer

$(1, 9/4)$

Vedclass · Answer

(D) Given $f(x) = ax^2 + bx + c$. Since the absolute term is $3$,$c = 3$. Since the leading coefficient is unity,$a = 1$. Thus,$f(x) = x^2 + bx + 3$.Since the parabola is symmetric about $x = 1$,the vertex is at $x = 1$. The derivative $f'(x) = 2x + b$. Setting $f'(1) = 0$,we get $2(1) + b = 0$,so $b = -2$.Therefore,the polynomial is $f(x) = x^2 - 2x + 3$.We can rewrite this as $y = (x - 1)^2 + 2$,or $(x - 1)^2 = 1(y - 2)$.Comparing this with the standard form of a parabola $(x - h)^2 = 4a(y - k)$,we have $h = 1$,$k = 2$,and $4a = 1$,which means $a = 1/4$.The focus of the parabola $(x - h)^2 = 4a(y - k)$ is given by $(h, k + a)$.Substituting the values,the focus is $(1, 2 + 1/4) = (1, 9/4)$.

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