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find the equation of a parabola with a vertex of (2,8) and passes through the point (10,-4)?

Sagot :

Answer:

[tex]\displaystyle y = -\frac{3}{16}\, (x - 2)^{2} + 8[/tex].

Step-by-step explanation:

In general, if the vertex of a parabola is [tex](h,\, k)[/tex] (where [tex]h[/tex] and [tex]k[/tex] are constants,) the equation of that parabola would be [tex]y = a\, (x - h)^{2} + k[/tex] for some constant [tex]a[/tex] ([tex]a \ne 0[/tex].) This equation is the vertex form equation of this parabola.

In this question, it is given that the vertex of this parabola is [tex](2,\, 8)[/tex]. Thus, [tex]h = 2[/tex] and [tex]k = 8[/tex]. The equation of this parabola would be [tex]y = a\, (x - 2)^{2} + 8[/tex] for some constant [tex]a[/tex].

Finding the value of this constant [tex]a[/tex] requires the coordinates of a point on this parabola other than the vertex [tex](2,\, 8)[/tex].

Since [tex](10,\, -4)[/tex] is a point on this parabola, [tex]x = 10[/tex] and [tex]y = (-4)[/tex] should satisfy the equation of this parabola [tex]y = a\, (x - 2)^{2} + 8[/tex].

Thus:

[tex](-4) = a\, (10 - 2)^{2} + 8[/tex].

Solve this equation for [tex]a[/tex]:

[tex]\displaystyle a = -\frac{3}{16}[/tex].

Thus, the equation of this parabola in vertex form would be:

[tex]\displaystyle y = -\frac{3}{16}\, (x - 2)^{2} + 8[/tex].