Question

The point of extrema of the equation $y=37x^2+222x-51$ is:

• A. $(-3,-384)$
• B. $(-3,91)$
• C. $(3,-91)$
• D. $(3,273)$

Answer 1

The correct option is A $(-3,-384)$

Given: $y=37x^2+222x-51$
Since coefficient of $x^2\text{is}37>0$, we get upward facing parabola and its minimum value lies at vertex.
Here, $y=37x^2+222x-51$
Differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=37\left(2\right)x+222$
$\frac{dy}{dx}=74x+222$
For critical point or $x$-coordinate of vertex, we get $\frac{dy}{dx}=0$
$\Rightarrow 74x+222=0$
$\Rightarrow x=-3$
$\&y_{min}=37(-3{)}^2+222(-3)-51$
$\Rightarrow y_{min}=333-666-51=-384$
Hence $y_{min}=-384$
Thus, coordinates of the vertex of the quadratic expression is:
$(-3,-384)$