Question

If G(x) = x² − 5x + 5,

find G'(a) and use it to find equations of the tangent lines to the curve
y = x² − 5x + 5 at the points (0, 5)
G'(a)=?
y1(x)=?pass through (0,5)
y2(x)=? passthrough (4,1)

Answer 1

To find G'(a), take the derivative of G(x) using the power rule. Use G'(a) to find equations of tangent lines passing through given points.

Step-by-step explanation:

To find G'(a), we need to take the derivative of G(x). The derivative of a quadratic function is found by applying the power rule: for a term of the form ax^n, the derivative is nx^(n-1). Applying this rule to G(x), we have:

G'(x) = 2x - 5

To find G'(a), we substitute a in place of x:

G'(a) = 2a - 5

Now, let's use G'(a) to find the equations of the tangent lines to the curve y = x² - 5x + 5 at the points (0, 5) and (4, 1).

For the point (0, 5), we know that the slope of the tangent line is G'(0). Substituting 0 in place of a, we have:

G'(0) = 2(0) - 5 = -5

So, the equation of the tangent line passing through (0, 5) is y = -5x + 5.

For the point (4, 1), we know that the slope of the tangent line is G'(4). Substituting 4 in place of a, we have:

G'(4) = 2(4) - 5 = 3

So, the equation of the tangent line passing through (4, 1) is y = 3x - 11.