Question

Determine the equation for the parabola graphed below?

Answer 1

Answer:

y = $\frac{1}{2}$ x² - 2x + 1

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

From the graph (h, k) = (2, - 1), thus

y = a(x - 2)² - 1

To find a substitute a point on the curve (0, 1) into the equation

1 = 4a - 1 ( add 1 to both sides )

4a = 2 ( divide both sides by 4 )

a = $\frac{2}{4}$ = $\frac{1}{2}$, hence

y = $\frac{1}{2}$(x - 2)² - 1 ← in vertex form

Expand the factor and simplify

y = $\frac{1}{2}$(x² - 4x + 4) - 1

  = $\frac{1}{2}$ x² - 2x + 2 - 1

  = $\frac{1}{2}$ x² - 2x + 1 ← in standard form

Answer 2

Answer:

The equation of parabola is:

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

Step-by-step explanation:

We know that the vertex form of the equation of parabola is given by:

                 $y=a(x-h)^2+k$

where the vertex of the parabola is (h,k).

Now, from the graph i.e. provided to us we see that the vertex of the parabola is located at (2,-1)

i.e.

(h,k)=(2,-1)

i.e.

h=2 and k= -1

Hence, we have the equation of the parabola as:

$y=a(x-2)^2+(-1)\\\\i.e.\\\\y=a(x-2)^2-1$

Now, with the help of a passing through point of the parabola we may easily obtain the value of a.

The parabola passes through (0,1)

Hence, on putting x=0 and y=1 we have:

$1=a(0-2)^2-1\\\\i.e.\\\\1=a\times 4-1\\\\i.e.\\\\4a-1=1\\\\i.e.\\\\4a=1+1\\\\i.e.\\\\4a=2\\\\i.e.\\\\a=\dfrac{2}{4}\\\\i.e.\\\\a=\dfrac{1}{2}$

Hence, the equation of parabola will be:

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

On expanding the square term we have:

$y=\dfrac{1}{2}(x^2+(-2)^2-2\times x\times 2)-1\\\\i.e.\\\\y=\dfrac{1}{2}(x^2+4-4x)-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2+\dfrac{1}{2}\times 4+\dfrac{1}{2}\times (-4x)-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2+2-2x-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2-2x+2-1\\\\i.e.\\\\y=\dfrac{1}{2}x^2-2x+1$