Question Video: Finding Unknown Coefficients in a Polynomial Function given the Equation of Its Tangent at a Given Point | Nagwa Question Video: Finding Unknown Coefficients in a Polynomial Function given the Equation of Its Tangent at a Given Point | Nagwa

# Question Video: Finding Unknown Coefficients in a Polynomial Function given the Equation of Its Tangent at a Given Point Mathematics • Second Year of Secondary School

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The line π₯ β π¦ β 3 = 0 touches the curve π¦ = ππ₯Β³ + ππ₯Β² at (1, β2). Find π and π.

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### Video Transcript

The line π₯ minus π¦ minus three equals zero touches the curve π¦ equals ππ₯ cubed plus ππ₯ squared at one, negative two. Find π and π.

The key information given in this question is that the line and the curve touch at this point with coordinates one, negative two. But the line does not cross the curve, which means that the line π₯ minus π¦ minus three equals zero is a tangent to the given curve at this point. We know that the gradient of a curve is equal to the gradient of the tangent to the curve at that point. The equation of our line is π₯ minus π¦ minus three equals zero. And when rearranging, we see that this is equivalent to π¦ equals π₯ minus three. Comparing with π¦ equals ππ₯ plus π, thatβs the general form of the equation of a straight line in slope-intercept form, we see that the slope of our tangent is one. Can we find an expression for the slope of the curve? Well, we can do this by differentiation. By applying the power rule, we see that dπ¦ by dπ₯ is equal to three ππ₯ squared plus two ππ₯.

Next, we evaluate this gradient function at the point one, negative, two. So we substitute π₯ equals one into our gradient function, giving three π plus two π. We can then equate the gradient of the curve at this point with the gradient of the tangent to the curve at this point. And it gives an equation involving π and π: three π plus two π is equal to one. We canβt solve this equation because we have only one equation and two unknowns. So weβre going to need to find a second equation.

The point one, negative two lies on both the curve and the tangent. So if we substitute the values of one and negative two into the equation of the curve, weβll get a second equation connecting π and π. We have π multiplied by one cubed plus π multiplied by one squared is equal to negative two, simplifying to π and π equals negative two. We now have two linear equations in π and π, which we need to solve simultaneously. We can multiply equation two by two, as this will make the coefficient of π the same as it is in equation one.

Weβll then subtract the second equation from the first to eliminate the π terms, giving π equals five. Substituting this value for π into our original equation two thatβs π and π equals negative two gives five plus π equals negative two. And by subtracting five, we see that π is equal to negative seven. So we found the values of π and π. π is equal to five and π is equal to negative seven.

A reminder is that the key point that we used in this question is that the slope of a curve is equal to the slope of the tangent to the curve at that point.

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