On the curve y = x²+3x-4 find the tangent parallel
On the curve $y = x^{2}+3x-4$ find the tangent parallel to each of the lines 1° $y = x$; 2° $y = 3x-3$; 3° $2x-3y+1 = 0$.
Explanation of Method
Parallel means equal slope, and the slope fixes the point.
Step 1: The general slope
\[ y' = 2x+3, \qquad k\bigl(x_{0}\bigr) = 2x_{0}+3, \qquad y_{0} = x_{0}^{2}+3x_{0}-4. \tag{1} \]
Requiring $k\bigl(x_{0}\bigr)$ to equal the given slope determines $x_{0}$, and with it the point of tangency.
Step 2: 1° — slope $1$
\[ 2x_{0}+3 = 1 \;\Longrightarrow\; x_{0} = -1, \qquad y_{0} = 1-3-4 = -6, \]
\[ y+6 = 1\cdot\bigl(x+1\bigr) \;\Longrightarrow\; y = x-5. \]
Step 3: 2° — slope $3$
\[ 2x_{0}+3 = 3 \;\Longrightarrow\; x_{0} = 0, \qquad y_{0} = -4, \qquad y = 3x-4. \]
Step 4: 3° — slope $\tfrac23$
The line $2x-3y+1 = 0$ is $y = \tfrac23x+\tfrac13$, of slope $\tfrac23$; hence
\[ 2x_{0}+3 = \frac23 \;\Longrightarrow\; 2x_{0} = -\frac73 \;\Longrightarrow\; x_{0} = -\frac76, \]
\[ y_{0} = \frac{49}{36}-\frac{7}{2}-4 = \frac{49-126-144}{36} = -\frac{221}{36}, \]
\[ y+\frac{221}{36} = \frac23\left(x+\frac76\right) \;\Longrightarrow\; y = \frac23x+\frac{28}{36}-\frac{221}{36} = \frac23x-\frac{193}{36}. \]
Step 5: Verification
\[y'(-1) = 1,\quad -1-5 = -6 \;\;\checkmark; \qquad y'(0) = 3,\quad -4 \;\;\checkmark\]
\[y'\!\left(-\tfrac76\right) = -\tfrac73+3 = \tfrac23 \;\;\checkmark\]
\[\frac23\left(-\frac76\right)-\frac{193}{36} = -\frac{28}{36}-\frac{193}{36} = -\frac{221}{36} = y_{0} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,1^{\circ}\ y = x-5 \ \text{ at } \bigl(-1,-6\bigr); \qquad 2^{\circ}\ y = 3x-4 \ \text{ at } \bigl(0,-4\bigr)\,}\]
Final answer:
\[\boxed{\,3^{\circ}\ y = \frac23x-\frac{193}{36} \ \text{ at } \left(-\frac76,\,-\frac{221}{36}\right)\,}\]
Remark
Remark. A parabola has exactly one tangent of each prescribed slope, because $y' = 2x+3$ is a bijection of the line onto itself. That is special to the quadratic: for $y = x^{3}$, say, each non-negative slope occurs twice and each negative slope not at all.