
How do you write the vertex form equation of each parabola given vertex $\left( {8, - 1} \right)$, y-intercept: -17?
Answer
571.2k+ views
Hint: Analyze the given. Substitute the coordinates of the vertex in the general equation of the horizontal and vertical parabola, and you’ll get an equation. In that equation, substitute the coordinates of the y-intercepts in which the vertex passes through and find out the value of a. Now, try to get the final answer.
Complete step-by-step solution:
As you know the general equation of a horizontal parabola is:
${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$
As given vertex to us is $\left( {8, - 1} \right)$, therefore we have,
$ \Rightarrow {\left( {y - \left( { - 1} \right)} \right)^2} = 4a\left( {x - 8} \right)$
Simplify the equation,
$ \Rightarrow {\left( {y + 1} \right)^2} = 4a\left( {x - 8} \right)$............….. (1)
Now substitute the y-intercept in the above equation,
$ \Rightarrow {\left( { - 17 + 1} \right)^2} = 4a\left( {0 - 8} \right)$
Simplify the terms,
$ \Rightarrow {\left( { - 16} \right)^2} = - 32a$
Divide both sides by -32,
$ \Rightarrow a = \dfrac{{256}}{{ - 32}}$
Divide numerator by the denominator,
$ \Rightarrow a = - 8$
Substitute the values in equation (1),
$ \Rightarrow {\left( {y + 1} \right)^2} = 4 \times - 8 \times \left( {x - 8} \right)$
Simplify the terms,
$ \Rightarrow {\left( {y + 1} \right)^2} = - 32\left( {x - 8} \right)$
As you know the general equation of a vertical parabola is:
${\left( {x - h} \right)^2} = 4a\left( {y - k} \right)$
As given vertex to us is $\left( {8, - 1} \right)$, therefore we have,
$ \Rightarrow {\left( {x - 8} \right)^2} = 4a\left( {y - \left( { - 1} \right)} \right)$
Simplify the equation,
$ \Rightarrow {\left( {x - 8} \right)^2} = 4a\left( {y + 1} \right)$..............….. (2)
Now substitute the y-intercept in the above equation,
$ \Rightarrow {\left( {0 - 8} \right)^2} = 4a\left( { - 17 + 1} \right)$
Simplify the terms,
$ \Rightarrow {\left( { - 8} \right)^2} = - 64a$
Divide both sides by -64,
$ \Rightarrow a = \dfrac{{64}}{{ - 64}}$
Divide numerator by the denominator,
$ \Rightarrow a = - 1$
Substitute the values in equation (2),
$ \Rightarrow {\left( {x - 8} \right)^2} = 4 \times - 1 \times \left( {y + 1} \right)$
Simplify the terms,
$ \Rightarrow {\left( {x - 8} \right)^2} = - 4\left( {y + 1} \right)$
Hence, the vertex form equation of a horizontal parabola is ${\left( {y + 1} \right)^2} = - 32\left( {x - 8} \right)$ and of the vertical parabola is ${\left( {x - 8} \right)^2} = - 4\left( {y + 1} \right)$.
Note: For these kinds of questions we must remember the general equation of a horizontal parabola and then put vertex's coordinates in it to get the equation in only one variable 'a'. Now, Put the coordinates of the point given to be on the parabola and solve to get the value of 'a'. Put the value of 'a' in the equation to get the desired equation.
Complete step-by-step solution:
As you know the general equation of a horizontal parabola is:
${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$
As given vertex to us is $\left( {8, - 1} \right)$, therefore we have,
$ \Rightarrow {\left( {y - \left( { - 1} \right)} \right)^2} = 4a\left( {x - 8} \right)$
Simplify the equation,
$ \Rightarrow {\left( {y + 1} \right)^2} = 4a\left( {x - 8} \right)$............….. (1)
Now substitute the y-intercept in the above equation,
$ \Rightarrow {\left( { - 17 + 1} \right)^2} = 4a\left( {0 - 8} \right)$
Simplify the terms,
$ \Rightarrow {\left( { - 16} \right)^2} = - 32a$
Divide both sides by -32,
$ \Rightarrow a = \dfrac{{256}}{{ - 32}}$
Divide numerator by the denominator,
$ \Rightarrow a = - 8$
Substitute the values in equation (1),
$ \Rightarrow {\left( {y + 1} \right)^2} = 4 \times - 8 \times \left( {x - 8} \right)$
Simplify the terms,
$ \Rightarrow {\left( {y + 1} \right)^2} = - 32\left( {x - 8} \right)$
As you know the general equation of a vertical parabola is:
${\left( {x - h} \right)^2} = 4a\left( {y - k} \right)$
As given vertex to us is $\left( {8, - 1} \right)$, therefore we have,
$ \Rightarrow {\left( {x - 8} \right)^2} = 4a\left( {y - \left( { - 1} \right)} \right)$
Simplify the equation,
$ \Rightarrow {\left( {x - 8} \right)^2} = 4a\left( {y + 1} \right)$..............….. (2)
Now substitute the y-intercept in the above equation,
$ \Rightarrow {\left( {0 - 8} \right)^2} = 4a\left( { - 17 + 1} \right)$
Simplify the terms,
$ \Rightarrow {\left( { - 8} \right)^2} = - 64a$
Divide both sides by -64,
$ \Rightarrow a = \dfrac{{64}}{{ - 64}}$
Divide numerator by the denominator,
$ \Rightarrow a = - 1$
Substitute the values in equation (2),
$ \Rightarrow {\left( {x - 8} \right)^2} = 4 \times - 1 \times \left( {y + 1} \right)$
Simplify the terms,
$ \Rightarrow {\left( {x - 8} \right)^2} = - 4\left( {y + 1} \right)$
Hence, the vertex form equation of a horizontal parabola is ${\left( {y + 1} \right)^2} = - 32\left( {x - 8} \right)$ and of the vertical parabola is ${\left( {x - 8} \right)^2} = - 4\left( {y + 1} \right)$.
Note: For these kinds of questions we must remember the general equation of a horizontal parabola and then put vertex's coordinates in it to get the equation in only one variable 'a'. Now, Put the coordinates of the point given to be on the parabola and solve to get the value of 'a'. Put the value of 'a' in the equation to get the desired equation.
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