Question

A quadratic curve has a minimum point P(1, –12). Given that the distance between the minimum point and its x-intercepts is 13 units, find the equation of this quadratic curve.

Answer 1

Answer:

$y=\dfrac{12}{25}(x-1)^2-12$

Step-by-step explanation:

We are told that the distance between the minimum point and the x-intercepts is 13 units.  Therefore, we can use the distance between two points formula to find the x-values of the x-intercepts.

As the x-intercepts are the points at which the curve crosses the x-axis, the y-value of the x-intercepts is always zero.

Distance between two points

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points.}$

Given:

• d = 13
• Let (x₁, y₁) = minimum point = (1, -12)
• Let (x₂, y₂) = x-intercept = (x, 0)

Substitute the given values into the formula and solve for x:

$\implies \sqrt{(x-1)^2+(0-(-12))^2}=13$

$\implies \sqrt{(x-1)^2+12^2}=13$

$\implies \sqrt{x^2-2x+1+144}=13$

$\implies \sqrt{x^2-2x+145}=13$

$\implies x^2-2x+145=169$

$\implies x^2-2x-24=0$

$\implies x^2-6x+4x-24=0$

$\implies x(x-6)+4(x-6)=0$

$\implies (x+4)(x-6)=0$

$\implies (x+4)=0 \implies x=-4$

$\implies (x-6)=0 \implies x=6$

Therefore, the x-intercepts of the quadratic curve are:

• (-4, 0) and (6, 0)

Vertex form of a quadratic equation

$y=a(x-h)^2+k$

where:

• (h, k) is the vertex (minimum/maximum point).
• a is some constant.

To find a, substitute the vertex (1, -12) and one of the x-intercepts (6, 0) into the formula:

$\implies 0=a(6-1)^2-12$

$\implies 0=25a-12$

$\implies 25a=12$

$\implies a=\dfrac{12}{25}$

Substitute the vertex and the found value of a into the vertex formula to find the equation of the quadratic curve:

$\implies y=\dfrac{12}{25}(x-1)^2-12$

Learn more about vertex form here:

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