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(i) Express y = x² – 4x + 7 in the form (x – a)² + b, hence, write down the coordinates of the minimum point on the curve。

(ii) Find the two values of m for which the line y = mx – 3m is a tangent to the curve y = x² – 4x + 7.
For each value of m, find the coordinates of the point where the line touches the curve.

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Final answer:

(i) To express the equation y = x² – 4x + 7 in the form (x – a)² + b, we need to complete the square. The equation can be written as (x – 2)² + 3, and the coordinates of the minimum point on the curve are (2, 3). (ii) The value of m for which the line y = mx – 3m is a tangent to the curve y = x² – 4x + 7 is 1. However, there are no points where the line touches the curve for m = 1.

Step-by-step explanation:

(i) To express the equation y = x² – 4x + 7 in the form (x – a)² + b, we need to complete the square. First, let's rearrange the equation:

y = (x² – 4x) + 7

Now, let's complete the square for the quadratic term (x² – 4x):

y = (x² – 4x + 4) + 7 – 4

Next, simplify the equation:

y = (x – 2)² + 3

Therefore, the equation y = x² – 4x + 7 can be written in the form (x – 2)² + 3. The coordinates of the minimum point on the curve are (2, 3).

(ii) To find the values of m for which the line y = mx – 3m is a tangent to the curve y = x² – 4x + 7, we need to equate the equations and solve for x. Setting the equations equal to each other:

(x – 2)² + 3 = mx – 3m

Simplifying the equation:

x² – 4x + 4 + 3 = mx – 3m

Combining like terms:

x² – 4x + 7 = mx – 3m

From here, we can equate the coefficients of x on both sides:

1 = m

Therefore, the value of m for which the line is a tangent to the curve is 1. Substituting this value into the equation y = mx – 3m, we can find the coordinates of the point where the line touches the curve. For m = 1:

y = x – 3

Setting the equation equal to y = x² – 4x + 7:

x² – 4x + 7 = x – 3

Simplifying the equation:

x² – 5x + 10 = 0

Using the quadratic formula to solve for x:

x = (-(-5) ± √((-5)² – 4(1)(10))) / 2(1)

Simplifying the expression:

x = (5 ± √(25 – 40)) / 2

Since the discriminant is negative, there are no real solutions for x. Therefore, there are no points where the line touches the curve for m = 1.

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