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7. The vertex of a parabola is (5,-3) and another point on the parabola is (1,5). ** a. Write the equation for this parabola. b. Which of the following points is also on the parabola? (Show your work, of course) a) (0, 3) b) (-1, 9) c) (-1, 15) d) (7, 7)

Sagot :

The vertex form of the parabola is

[tex]y=a(x-h)^2+k[/tex]

Where (h, k) are the coordinates of its vertex

Since the vertex of the parabola is (5, -3), then

h = 5 and k = -3

Substitute them in the form above

[tex]\begin{gathered} y=a(x-5)^2+(-3) \\ y=a(x-5)^2-3 \end{gathered}[/tex]

To find the value of (a) we will use the given point (1, 5)

Substitute x by 1 and y by 5

[tex]\begin{gathered} 5=a(1-5)^2-3 \\ 5=a(-4)^2-3 \\ 5=16a-3 \end{gathered}[/tex]

Add both sides by 3

[tex]\begin{gathered} 5+3=16a-3+3 \\ 8=16a \end{gathered}[/tex]

Divide both sides by 16 to find (a)

[tex]\begin{gathered} \frac{8}{16}=\frac{16a}{16} \\ \frac{1}{2}=a \\ a=\frac{1}{2} \end{gathered}[/tex]

Substitute a by 1/2 in the equation above, then

a. The equation of the parabola is

[tex]y=\frac{1}{2}(x-5)^2-3[/tex]

Let us substitute x and y in the equation by each answer

Since x = 0 and y = 3

[tex]\begin{gathered} 3=\frac{1}{2}(0-5)^2-3 \\ 3=12.5-3 \\ 3=9.5 \\ \text{LHS}\ne RHS \end{gathered}[/tex]

Since x = -1 and y = 9

[tex]\begin{gathered} 9=\frac{1}{2}(-1-5)^2-3 \\ 9=\frac{1}{2}(-6)^2-3 \\ 9=\frac{1}{3}(36)-3 \\ 9=12-3 \\ 9=9 \\ \text{LHS}=\text{RHS} \end{gathered}[/tex]

The point (-1, 9) lies on the parabola

The answer is b

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