Find the point on the curve `y=3x^2-4x+5` where the tangent line is parallel to the line y=-22x+7

Options

A. (0,5)
B. (-1,12)
C. (2,25)
D. (-3,34)

Answer

To find the point on the curve \( y = 3x^2 - 4x + 5 \) where the tangent line is parallel to the line \( y = -22x + 7 \), we will follow these steps:

### Step 1: Identify the slope of the given line
The equation of the line is given in the slope-intercept form \( y = mx + c \), where \( m \) is the slope. From the equation \( y = -22x + 7 \), we can see that the slope \( m = -22 \).

### Step 2: Find the derivative of the curve
To find the slope of the tangent line to the curve at any point \( x \), we need to compute the derivative of the function \( y = 3x^2 - 4x + 5 \).

\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - 4x + 5) = 6x - 4 \]

### Step 3: Set the derivative equal to the slope of the line
Since we want the tangent line to be parallel to the line \( y = -22x + 7 \), we set the derivative equal to the slope of the line:

\[ 6x - 4 = -22 \]

### Step 4: Solve for \( x \)
Now, we will solve the equation for \( x \):

\[ 6x - 4 = -22 \\ 6x = -22 + 4 \\ 6x = -18 \\ x = \frac{-18}{6} = -3 \]

### Step 5: Find the corresponding \( y \) value
Now that we have \( x = -3 \), we can find the corresponding \( y \) value by substituting \( x \) back into the original curve equation:

\[ y = 3(-3)^2 - 4(-3) + 5 \\ y = 3(9) + 12 + 5 \\ y = 27 + 12 + 5 \\ y = 44 \]

### Step 6: Write the point
The point on the curve where the tangent line is parallel to the given line is:

\[ (-3, 44) \]

### Final Answer
The point on the curve \( y = 3x^2 - 4x + 5 \) where the tangent line is parallel to the line \( y = -22x + 7 \) is \( (-3, 44) \).

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