13 votes

If $y=5x^2+3$, then the tangent at $x=0$, $y=3$

- passes through $x=0,y=0$
- has a slope of $+1$
- is parallel to the $x$-axis
- has a slope of $-1$

15 votes

Best answer

Given, $y=5x^2+3$

Slope $dy/dx = 10x$

Given point $x=0,y=3: (0,3)$

Now, at $(0,3)$ we have the slope as $dy/dx = 10 \times 0=0,$ which means no matter how much ever we change 'x',change in y is zero or y is a constant (y=3,here) i.e line is parallel to $X$- axis.

Options B and D eliminated as they have slope as '+1' and '-1'.

Option A is eliminated as given the tangent is at point $(0,3)$ and we found $slope=0$, which means it is parallel to $X$ axis. The equation of line is nothing but $y=3$, which never passes through $(0,0).$ Hence A too eliminated.

Answer: Option $C)$

Slope $dy/dx = 10x$

Given point $x=0,y=3: (0,3)$

Now, at $(0,3)$ we have the slope as $dy/dx = 10 \times 0=0,$ which means no matter how much ever we change 'x',change in y is zero or y is a constant (y=3,here) i.e line is parallel to $X$- axis.

Options B and D eliminated as they have slope as '+1' and '-1'.

Option A is eliminated as given the tangent is at point $(0,3)$ and we found $slope=0$, which means it is parallel to $X$ axis. The equation of line is nothing but $y=3$, which never passes through $(0,0).$ Hence A too eliminated.

Answer: Option $C)$