Question Statement

Find the coordinates of the vertex of each parabola by differentiating its equation on both sides and then solving for the horizontal (or vertical) tangent.

(i) $y=x^2-4x+10$

(ii) $4x^2+24x+39-3y=0$

(iii) $y^2-10y+4x+28=0$

(iv) $y^2-14y=-3x-45y$

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Background and Explanation

The vertex of a parabola is the point where the curve turns. At this specific point, the tangent line to the parabola is either horizontal (slope $\frac{dy}{dx}=0$) if the parabola opens upwards or downwards, or vertical (slope $\frac{dy}{dx}=\infty$) if the parabola opens to the left or right. By finding where these conditions occur, we can identify the coordinates of the vertex.

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Solution

Part (i)

Given equation: $y=x^2-4x+10$

1. Differentiate with respect to $x$: $\frac{dy}{dx}=2x-4$ This expression represents the slope of the tangent line.

2. Check for horizontal tangent: Set the slope to zero: $\frac{dy}{dx}=0⟹2x-4=0⟹x=2$

3. Find the corresponding $y$-coordinate: Substitute $x=2$ into the original equation: $\begin{array}{rl}y& =(2{)}^2-4(2)+10\\ y& =4-8+10\\ y& =6\end{array}$

4. Check for vertical tangent: Set the slope to infinity: $\frac{dy}{dx}=\infty ⟹2x-4=\infty ⟹x=\infty \text{ (not possible)}$

Result: The coordinates of the vertex are $(2,6)$.

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Part (ii)

Given equation: $4x^2+24x+39-3y=0$

1. Differentiate with respect to $x$: $\begin{array}{rl}8x+24-3\frac{dy}{dx}& =0\\ 3\frac{dy}{dx}& =8x+24\\ \frac{dy}{dx}& =\frac{8}{3}(x+3)\end{array}$

2. Check for horizontal tangent: Set $\frac{dy}{dx}=0$: $\frac{8}{3}(x+3)=0⟹x+3=0⟹x=-3$

3. Find the corresponding $y$-coordinate: Substitute $x=-3$ into the original equation: $\begin{array}{rl}4(-3{)}^2+24(-3)+39-3y& =0\\ 36-72+39& =3y\\ 3& =3y\\ y& =1\end{array}$

4. Check for vertical tangent: $\frac{dy}{dx}=\infty ⟹\frac{8}{3}(x+3)=\infty ⟹x=\infty \text{ (not possible)}$

Result: The coordinates of the vertex are $(-3,1)$.

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Part (iii)

Given equation: $y^2-10y+4x+28=0$

1. Differentiate with respect to $x$: $\begin{array}{rl}2y\frac{dy}{dx}-10\frac{dy}{dx}+4& =0\\ (2y-10)\frac{dy}{dx}& =-4\\ \frac{dy}{dx}& =\frac{-4}{2y-10}=\frac{2}{5-y}\end{array}$

2. Check for horizontal tangent: Set $\frac{dy}{dx}=0$: $\frac{2}{5-y}=0⟹2=0\text{ (not possible)}$

3. Check for vertical tangent: A vertical tangent occurs when the denominator of the slope is zero: $\frac{dy}{dx}=\infty ⟹5-y=0⟹y=5$

4. Find the corresponding $x$-coordinate: Substitute $y=5$ into the original equation: $\begin{array}{rl}(5{)}^2-10(5)+4x+28& =0\\ 25-50+4x+28& =0\\ 4x+3& =0\\ x& =-\frac{3}{4}\end{array}$

Result: The coordinates of the vertex are $\left(-\frac{3}{4},5\right)$.

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Part (iv)

Given equation: $y^2-14y=-3x-45y$

1. Differentiate with respect to $x$: $\begin{array}{rl}2y\frac{dy}{dx}-14\frac{dy}{dx}& =-3-45\frac{dy}{dx}\\ (2y-14+45)\frac{dy}{dx}& =-3\\ (2y+31)\frac{dy}{dx}& =-3\\ \frac{dy}{dx}& =\frac{-3}{2y+31}\end{array}$

2. Check for horizontal tangent: Set $\frac{dy}{dx}=0$: $\frac{-3}{2y+31}=0⟹-3=0\text{ (not possible)}$

3. Check for vertical tangent: A vertical tangent occurs when the denominator is zero: $\frac{dy}{dx}=\infty ⟹2y+31=0⟹y=-\frac{31}{2}$

4. Find the corresponding $x$-coordinate: Substitute $y=-\frac{31}{2}$ into the original equation: $\begin{array}{rl}{\left(-\frac{31}{2}\right)}^2-14\left(-\frac{31}{2}\right)& =-3x-45\left(-\frac{31}{2}\right)\\ \frac{961}{4}+217& =-3x+\frac{1395}{2}\\ 3x& =\frac{1395}{2}-\frac{961}{4}-217\\ 3x& =\frac{2790-961-868}{4}\\ 3x& =\frac{961}{4}\\ x& =\frac{961}{12}\end{array}$

Result: The coordinates of the vertex are $\left(\frac{961}{12},-\frac{31}{2}\right)$.

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Key Formulas or Methods Used

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$, especially when $y$ is not isolated.
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$.
• Horizontal Tangent Condition: Set the first derivative $\frac{dy}{dx}=0$.
• Vertical Tangent Condition: Set the denominator of the first derivative $\frac{dy}{dx}$ to $0$ (representing an undefined or infinite slope).

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Summary of Steps

1. Differentiate the given equation of the parabola with respect to $x$ to find the expression for $\frac{dy}{dx}$.
2. Determine if the parabola has a horizontal or vertical tangent at the vertex.
3. For a horizontal tangent, set the numerator of $\frac{dy}{dx}$ to zero and solve for the variable.
4. For a vertical tangent, set the denominator of $\frac{dy}{dx}$ to zero and solve for the variable.
5. Substitute the resulting value back into the original equation to find the second coordinate of the vertex.