Find the abscissa of the point on the graph of the function y = x ^ 2-2x + 5, in which the tangent

Find the abscissa of the point on the graph of the function y = x ^ 2-2x + 5, in which the tangent to it is parallel to the line y-2x = 0.

1. Let’s find the derivative of the function:

y = x ^ 2 – 2x + 5;
y ‘= 2x – 2.
2. Determine the slope of the straight line:

y – 2x = 0;
y = 2x;
k = 2.
3. The value of the derivative of the function at the point of tangency is equal to the slope of the tangent:

y ‘= k;
2×0 – 2 = 2;
2×0 = 2 + 2;
2×0 = 4;
x0 = 2.
4. We also calculate the ordinate of the point of tangency and compose the equation of the tangent:

y0 = x0 ^ 2 – 2×0 + 5;
y0 = 2 ^ 2 – 2 * 2 + 5 = 4 – 4 + 5 = 5.
Tangent equation:

y – y0 = y ‘(x0) (x – x0);
y – 5 = 2 (x – 2);
y – 5 = 2x – 4;
y = 2x – 4 + 5;
y = 2x + 1.
Answer. Touch point abscissa: 2.