The normal to the curve $y=2x^2-x+3$ at the point $P(1,4)$ meets at the curve again at the point $Q$. What is the $x$– coordinate of $Q$?
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$(-4/7), -1, 0$ are wrong answers. Accept them if you find anything!
[Math] The normal to the curve $y=2x^2-x+3$ at the point $P(1,4)$ meets at the curve again at the point $Q$. What is the $x$- coordinate of $Q$
calculusderivatives
Best Answer
The derivative is $4x-1$. Therefore the derivate at that point is 3. From that, the normal slope is $-\frac{1}{3}$ We can form a line from this: $y-4=-\frac{1}{3}(x-1)$ This simplifies to $y=-\frac{x}{3}+\frac{13}{3}$. So we need $$2x^2-x+3=-\frac{x}{3}+\frac{13}{3}$$ From this:$$2x^2-\frac{2x}{3}-\frac{4}{3}$$After multiplying by 3, which preserves the roots:$$6x^2-2x-4$$ Which factors to, $(6x+4)(x-1)$. This makes sense because the one root is at one, the point, and the other is at $x=-\frac{4}{6}=-\frac{2}{3}$. This is your answer.