The tangent to the curve y = x² - 5x + 5, parallel to the line 2y = 4x + 1, also passes through the

Q: Wait, why do we equate the derivative to the slope of the line? Isn't the derivative just a function?

That is a fantastic question! Think of the derivative $\frac{dy}{dx}$ not just as a function, but as a 'slope-finding machine.' For any value of $x$ you plug in, it tells you the steepness of the curve at that exact spot. Since we want a tangent that is parallel to our reference line, we are essentially asking the machine: 'At what $x$ is the steepness of this curve exactly equal to the slope of that line?' By setting them equal, we are finding the precise location where the curve and the line share the same direction.

Q: I found the point of tangency, but how do I know which option is correct? Do I have to check all of them?

You don't necessarily have to check all of them if you are confident in your tangent equation! Once you have the equation of the tangent line, you can simply plug the $x$-coordinate of each option into your equation. If the resulting $y$-value matches the $y$-coordinate of the option, that point lies on the line. It is a quick and reliable way to verify your work.

Q: Can I use the discriminant method (setting $D=0$ for the intersection of the line and parabola) instead of calculus?

Yes, you absolutely can! If you substitute the expression for $y$ from the line into the parabola equation, you get a quadratic in $x$. Since the line is a tangent, it must intersect the parabola at exactly one point, meaning the discriminant of that quadratic must be zero. Both methods are mathematically sound and elegant. Calculus is often faster, but the discriminant method is a powerful algebraic alternative!

Question

The tangent to the curve $y = x^2 - 5x + 5$, parallel to the line $2y = 4x + 1$, also passes through the point.

Options:

(A) $(1/4, 7/2)$

(B) $(7/2, 1/4)$

(C) $(1/8, 7)$

(D) $(1/8, -7)$

Sigma Percentile · Accepted Answer

### The Geometry of Parallelism: A Journey into Tangents

Imagine you are standing on a path defined by the curve $y = x^2 - 5x + 5$. It is a beautiful, sweeping parabola, opening upwards, inviting you to explore its slopes.

Now, imagine a straight road nearby, represented by the line $2y = 4x + 1$. Your mission is to find a tangent to this parabola that runs perfectly parallel to that road. This is not just a math problem; it is a study of alignment and harmony.

### Phase 1: Extracting the Slope

Before we dive into the calculus, we must understand our reference. The line $2y = 4x + 1$ is our compass.

To understand its direction, we need its slope. By rearranging the equation into the slope-intercept form $y = mx + c$, we get:

$$y = \frac{4x + 1}{2} = 2x + \frac{1}{2}$$

Here, the coefficient of $x$ is our slope, $m = 2$. Any line parallel to this one must share this exact same slope.

### Phase 2: The Power of Differentiation

Now, how do we find where our parabola mimics this slope? This is where the magic of calculus comes in.

The derivative $\frac{dy}{dx}$ is our tool for measuring the instantaneous steepness of the curve. Let us differentiate $y = x^2 - 5x + 5$ with respect to $x$:

$$\frac{dy}{dx} = 2x - 5$$

This expression is our slope-finding machine.

### Phase 3: The Intersection of Algebra and Geometry

We want the slope of the tangent to be $2$. So, we set our machine to this value:

$$2x - 5 = 2$$

Solving this linear equation, we add $5$ to both sides to get $2x = 7$, which leads us to $x = \frac{7}{2}$.

To find the corresponding $y$-coordinate, we substitute $x = \frac{7}{2}$ back into the original curve equation:

$$y = \left(\frac{7}{2}\right)^2 - 5\left(\frac{7}{2}\right) + 5$$

Calculating this, we get:

$$y = \frac{49}{4} - \frac{35}{2} + 5 = \frac{49 - 70 + 20}{4} = -\frac{1}{4}$$

Our point of tangency is $(\frac{7}{2}, -\frac{1}{4})$.

### Phase 4: Constructing the Tangent

We have the point $(\frac{7}{2}, -\frac{1}{4})$ and the slope $m = 2$. Using the point-slope form $y - y_1 = m(x - x_1)$, we write:

$$y - \left(-\frac{1}{4}\right) = 2\left(x - \frac{7}{2}\right)$$

Simplifying this, we get $y + \frac{1}{4} = 2x - 7$, which leads to:

$$y = 2x - \frac{29}{4}$$

### Phase 5: The Final Verification

Finally, we check which of the given options lies on this line. Testing the point $(\frac{1}{8}, -7)$, we substitute $x = \frac{1}{8}$ into our tangent equation:

$$y = 2\left(\frac{1}{8}\right) - \frac{29}{4} = \frac{1}{4} - \frac{29}{4} = -\frac{28}{4} = -7$$

The point satisfies the equation perfectly! We have successfully navigated the curve, found the parallel tangent, and verified our result. The final equation of the tangent is **$y = 2x - \frac{29}{4}$**.

Animated Solution for Mathematics - Differentiation: The tangent to the curve $y = x^{2} - 5 x + 5$ , parallel to the line $2 y = 4 x + 1$ , also passes through the point.

Select Answer:

Visualized Solution

Visualize the Curve and Reference Line

Extracting the Slope ( $m$ )

The Concept of Differentiation

Calculating the Derivative

Equating the Slopes

Solving for $x$

Finding the $y$ -coordinate

Calculating $y$

Equation of the Tangent

Simplifying the Equation

Verifying the Options

Conclusion & Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Parallelism

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the curve y=x2−5x+5, parallel to the line 2y=4x+1, also passes through the point.