At what point is the tangent to the parabola y=x^2-7x+3 parallel to the line 5x+y-3=0?

To find the point where the tangent to the parabola y=x^2 −7x + 3 is parallel to the line 5x+y−3=0, we need to determine both the slope of the tangent to the parabola and the slope of the given line. If two lines are parallel, their slopes are equal. First, let's find the slope of the given line 5x+y−3=0. We can rewrite this line in slope-intercept form (y=mx+b), where m is the slope: 5x+y−3=0 y=−5x+3 So, the slope of the given line is m=−5. Now, let's find the derivative of the parabola y=x^2 −7x+3 to determine the slope of the tangent line at any point on the parabola. y=x^2 −7x+3 Take the derivative with respect to x: dy /dx = 2x−7 Now, we want to find the x-value(s) where the slope of the tangent line is equal to -5 (the slope of the given line). So, we set the derivative equal to -5: 2x−7=−5 Now, solve for x: 2x=−5+7 2x=2 x=1 Now that we have found x=1, we can find the corresponding y-coordinate on the parabola: y=x^2 −7x+3 y=(1)^2 −7(1)+3 y=1−7+3 y=−3 So, at the point (1,−3), the tangent to the parabola y=x^2−7x+3 is parallel to the line 5x+y−3=0.