Answer: 16/81 (x-10)^2 -4
Step-by-step explanation:
To write a vertex equation with just a point and the vertex, you have to figure out the variables.
In vertex form, the equation is y = a (x-h)^2 + k
Your y is 12, x = 1, h = 10, and k = -4
Plug everything into equation
12 = a (1 - 10)^2 -4
12 = a (-9)^2 - 4
12 = 81a - 4
16 = 81a
16/81 = a
Now you know what the 'a' value is.
If you graph 16/81 (x-10)^2 -4 , you will get a point at (1,12) and a vertex of (10,-4)!
I hope this helps!
Solving a system of equations, it is found that the quadratic equation is defined by:
[tex]y = 5.0625x^2 - 101.25x + 108.1875[/tex]
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A quadratic equation has an equation in the following format:
[tex]y = ax^2 + bx + c[/tex]
[tex]V(-\frac{b}{2a}, -\frac{b^2 - 4ac}{4a})[/tex]
It passes through (1,12), which means that when [tex]x = 1, y = 12[/tex], thus:
[tex]12 = a + b + c[/tex]
[tex]a + b + c = 12[/tex]
The x-coordinate of the vertex is 10, thus:
[tex]-\frac{b}{2a} = 10[/tex]
[tex]b = -20a[/tex]
The y-coordinate of the vertex is -4, thus:
[tex]-\frac{b^2 - 4ac}{4a} = -4[/tex]
[tex]b^2 - 4ac = 16a[/tex]
Since [tex]b = -20a[/tex]
[tex](-20a)^2 - 4ac = 16a[/tex]
[tex]400a^2 - 16a - 4ac = 0[/tex]
[tex]a + b + c = 12[/tex]
[tex]a - 20a + c = 12[/tex]
[tex]c = 12 + 19a[/tex]
Then
[tex]400a^2 - 16a - 4a(12 + 19a) = 0[/tex]
[tex]324a^2 - 64a = 0[/tex]
[tex]a(324 - 64a) = 0[/tex]
Quadratic equation, thus [tex]a \neq 0[/tex]:
[tex]324 - 64a = 0[/tex]
[tex]a = \frac{324}{64}[/tex]
[tex]a = 5.0625[/tex]
[tex]b = -20a = -20(5.0625) = -101.25[/tex]
[tex]c = 12 + 19a = 12 + 19(5.0625) = 108.1875[/tex]
Thus, the quadratic equation is:
[tex]y = 5.0625x^2 - 101.25x + 108.1875[/tex]
A similar problem is given at https://brainly.com/question/6035046
