WAEC - Further Mathematics (2019 - No. 18)
Find the coordinates of the point in the curve y = 3x\(^2\) - 2x - 5 where the tangent is parallel to the line y = - 5 = 8x
\(\begin{pmatrix} - \frac{5}{3} &, 0 \end {pmatrix}\)
\(\begin{pmatrix} 0, & - \frac{5}{3} \end {pmatrix}\)
\(\begin{pmatrix} 0, & \frac{5}{3} \end {pmatrix}\)
\(\begin{pmatrix} \frac{5}{3} &, 0 \end {pmatrix}\)
คำอธิบาย
y = 8x + 5
m = 8
y = 3x\(^2\) - 2x - 5
\(\frac{dy}{dx}\) = 6x - 2x - 5
\(\frac{6x}{6} = \frac{10}{6}\)
x = \(\frac{5}{3}\)
y = 0