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f'(x) = 2x
Slope of tangent at (x = 1) = 2 * (1) = 2

The point (5, 3) lies on the tangent at x = 1.
The equation of the tangent at x = 1:
(y - 3)/(x - 5) = 2
2(x - 5) = y - 3
2x - 10 = y - 3
2x - y - 7 = 0

f(x) = y = x² + c
When x = 1, y = 1 + c
The point (1, 1 + c) also lies on 2x - y - 7 = 0
Substitute x = 1 and y = 1 + c into the equation 2x - y - 7 = 0:
2(1) - (1 + c) - 7 = 0
2 - 1 - c - 7 = 0
c = -6

Wouldn't this be precalculus?

f(x) = x² + c
f'(1) = 2(1) = 2

f(1) = 1 + c

Now we have two information: the tangent line has a gradient of 2 and passes through (1, 1 + c). We can now obtain its equation:

y - (1 + c) = 2(x - 1)
y- 1 - c = 2x - 2
y = 2x - 1 + c

From the question we know this equation is true for (5, 3). Hence substituting the values,

3 = 10 - 1 + c
c = -6

This graph illustrates the answer worked out below
https://www.wolframalpha.com/input/?i=y+%3D+x%5E2+-+6,+y+%3D+2x+-+7,+x+%3D+5,+(y+from+-7+to+6)

y = x^2 + c and when x = 1, y = c + 1
Slope of tangent line is given by
dy/dx = 2x which is 2 when x = 1
Line with slope 2 goes through (5, 3) and (1, c + 1)
(2 – c)/(5 – 1) = 2, so, c = -6

f(x) = x² + c ← this is a function

f(1) = 1 + c → the curve passes through the point (1 ; 1 + c)


f’(x) = 2x ← this is the derivative of the function

…but the derivative is too the slope of the tangent line to the curve at x

f’(1) = 2 ← this is the slope of the tangent line to the curve at: x = 1 derivative


The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

As the slope of the tangent line is (2), the equation of the tangent line becomes: y = 2x + b

The line tangent line passes through the point (1 ; 1 + c), so these coordinates must verify the tangent line.

y = 2x + b

b = y - 2x → you substitute x and y by the coordinates of the point (1 ; 1 + c)

b = (1 + c) - (2 * 1)

b = 1 + c - 2

b = c - 1

→ The equation of the tangent line is: y = 2x + (c - 1)

The tangent line passes through the point (5 ; 3), so these coordinates must verify the tangent line.

y = 2x + (c - 1) → you substitute x and y by the coordinates of the point (5 ; 3)

3 = 10 + c - 1

→ c = - 6

.

f(x) = x² + c
slope = f’(x) = 2x
∴ f’(1) = 2(1) = 2; thus the slope at x = 1 is 2

Equation of the tangent line of slope 2 that passes through the point (5, 3)
y - b = m(x - a) where (a, b) = (5, 3) and slope m = 2

y - 3 = 2(x - 5)
y = 2x - 7

f(x) = x² + c
When x = 1, f(1) = 1 + c
Point of intersection of the line y = 2x - 7 and the curve y = x² + c is ( 1, 1 + c)
y = 2x - 7
1 + c = 2(1) - 7
c = -6
━━━

first make the equation for tangent line at x=1,y=1*1+c
then substitute x=5,y=3 & solve for c.
c=-6