Question

The below quadratic function can model the natural shape of a banana. Now, we know that a parabolic shape must have a quadratic function, therefore an equation in standard form of f(x)=ax? + bx + c. To find an equation for the parabolic shape of the banana, we need to find the values of a, b, and c. From the banana picture above, we can see that a quadratic function is able to model the banana quite accurately, with a=0.1, b=0, and c=0. Therefore, the equation is f(x)=0.1x

Quadratic Function on a Banana

..0.1

y2O

(i) Name the shape of the banana curve from the above figure. (ii) Find the number of the zeroes of the polynomial for the shape of the banana. (iii)If the curve of banana represented by f(x) = x x- 12. Find its zeroes. (iv) If the representation of banana curves whose one zero is 4 and the sum of the zeroes is 0 then find the quadratic polynomial.​

Answer 1

Answer:

i) A parabola is formed with the curve of banana

ii) Since the line intersects on the x-axis.

Therefore number of zeroes is 1

iii) sum of Zeroes= 0

therefore, alpha + beta = 0

Step-by-step explanation:

For the Third part:-

f(x) = x²-x-12

= x²+3x-4x-12

=x(x+3)-4(x+3)

=(x-4)(x+3)

∴ x-4=0  or  x+3=0

∴ x=4 or x=-3

∴ α=4 and β= -3

therefore the other zero is -3

And both the zeroes are 4 and -3.

Answer 2

Concept:

Quadratic equations are algebraic expressions of the second degree. The Quadratic Formula is the most straightforward method for determining the roots of a quadratic equation. Certain quadratic equations are difficult to factor, thus we can utilise this quadratic formula to determine the roots as quickly as feasible.

Find:

(i) The shape of the banana.

(ii) The number of zeros of the polynomial of banana.

(iii) Find the zeros of the given polynomial.

(iv) Find the polynomial from the zeros.

Solution:

(i) The shape of the curve is a parabola.

(ii) From the figure, the curve touches the x-axis at only one point. The number of zeros is 1.

(iii)

$f(x)=x^2-x-12\\f(x)=x^2-4x+3x-12\\f(x)=x(x-4)+3(x-4)\\f(x)=(x-4)(x+3)$

The zeros of the polynomial are $-3, 4$.

(iv)

$\alpha=4\\\alpha + \beta=0$

$\beta=-4\\\alpha \beta=-16$

The general form of a quadratic equation:

$f(x)=x^2-(\alpha + \beta)x+\alpha \beta\\f(x)=x^2-0*x-16\\f(x)=x^2-16$

The quadratic equation is $x^2-16$.

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