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Part 1 - ApplicationPhysics shows that when an object is thrown upward with an initial velocity of v then its approximate height is given by this quadratic function.s = − 4. 9t^2+ 2.8t + 15I need help putting this into vertex form by completing the square.

Sagot :

EXPLANATION

Since we need to find the vertex of the parabola, we first should calculate the params (h,k), as shown as follows:

[tex]h=-\frac{b}{2a}\text{ k=f(h)}[/tex]

As the general form is as follows:

[tex]f(x)=ax^2+bx+c[/tex]

We have that a=-4.9, b=2.8 and c=15

Computing the value of h:

[tex]h=-\frac{2.8}{2\cdot(-4.9)}[/tex]

Multiplying numbers:

[tex]h=\frac{2.8}{9.8}=0.28571\ldots[/tex]

Now, we need to compute the value of k:

[tex]k=-4.9\cdot\: 0.28571\ldots^2+2.8\cdot\: 0.28571\ldots+15[/tex]

Multiplying numbers:

[tex]=-0.28571\ldots^2\cdot\: 4.9+0.8+15[/tex]

Adding numbers:

[tex]=15.8-0.28571\ldots^2\cdot\: 4.9[/tex][tex]=15.8-0.4[/tex][tex]\mathrm{Subtract\: the\: numbers\colon}\: 15.8-0.4=15.4[/tex][tex]=15.4[/tex][tex]k=15.4[/tex]

Therefore, the parabola vertex is (h,k) = (0.2857,15.4)

Thus, the equation of the parabola in vertex form is as follows:

[tex]f(x)=-4.9(x-0.2857)^2+15.4[/tex]