Solveeit Logo

Question

Question: How do identify the important parts of \(y = - 7{x^2}\) to graph it?...

How do identify the important parts of y=7x2y = - 7{x^2} to graph it?

Explanation

Solution

We have to find the properties of the given parabola. First rewrite the equation in vertex form. Next, use the vertex form of parabola, to determine the values of aa, hh, and kk. Next, find the vertex and the distance from the vertex to the focus. Then, find the focus, axis of symmetry and directrix. Use the properties of the parabola to analyse and graph the parabola. Select a few xx values, and plug them into the equation to find the corresponding yy values. The xx values should be selected around the vertex. Graph the parabola using its properties and the selected points.

Formula used:
Vertex form of a parabola: a(x+d)2+ea{\left( {x + d} \right)^2} + e
d=b2ad = \dfrac{b}{{2a}}
e=cb24ae = c - \dfrac{{{b^2}}}{{4a}}
Vertex form: y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k
Vertex: (h,k)\left( {h,k} \right)
p=14ap = \dfrac{1}{{4a}}
Focus: (h,k+p)\left( {h,k + p} \right)
Directrix: y=kpy = k - p

Complete step by step answer:
We have to find the properties of the given parabola.
So, first rewrite the equation in vertex form.
For this, complete the square for 7x2 - 7{x^2}.
Use the form ax2+bx+ca{x^2} + bx + c, to find the values of aa, bb, and cc.
a=7,b=0,c=0a = - 7,b = 0,c = 0
Consider the vertex form of a parabola.
a(x+d)2+ea{\left( {x + d} \right)^2} + e
Now, substitute the values of aa and bb into the formula d=b2ad = \dfrac{b}{{2a}}.
d=02(7)d = \dfrac{0}{{2\left( { - 7} \right)}}
Simplify the right side.
d=0\Rightarrow d = 0
Find the value of ee using the formula e=cb24ae = c - \dfrac{{{b^2}}}{{4a}}.
e=0024(7)e = 0 - \dfrac{{{0^2}}}{{4\left( { - 7} \right)}}
e=0\Rightarrow e = 0
Now, substitute the values of aa, dd, and ee into the vertex form a(x+d)2+ea{\left( {x + d} \right)^2} + e.
7x2- 7{x^2}
Set yy equal to the new right side.
y=7x2y = - 7{x^2}
Now, use the vertex form, y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k, to determine the values of aa, hh, and kk.
a=7a = - 7
h=0h = 0
k=0k = 0
Since the value of aa is negative, the parabola opens down.
Opens Down
Find the vertex (h,k)\left( {h,k} \right).
(0,0)\left( {0,0} \right)
Now, find pp, the distance from the vertex to the focus.
Find the distance from the vertex to a focus of the parabola by using the following formula.
14a\dfrac{1}{{4a}}
Substitute the value of aa into the formula.
14(7)\dfrac{1}{{4\left( { - 7} \right)}}
Multiply 44 by 7 - 7, we get
128\Rightarrow - \dfrac{1}{{28}}
Find the focus.
The focus of a parabola can be found by adding pp to the yy-coordinate kk if the parabola opens up or down.
(h,k+p)\left( {h,k + p} \right)
Now, substitute the known values of hh, pp, and kk into the formula and simplify.
(0,128)\left( {0, - \dfrac{1}{{28}}} \right)
Find the axis of symmetry by finding the line that passes through the vertex and the focus.
x=0x = 0
Find the directrix.
The directrix of a parabola is the horizontal line found by subtracting pp from the yy-coordinate kk of the vertex if the parabola opens up or down.
y=kpy = k - p
Now, substitute the known values of pp and kk into the formula and simplify.
y=128y = \dfrac{1}{{28}}
Use the properties of the parabola to analyse and graph the parabola.
Direction: Opens Down
Vertex: (0,0)\left( {0,0} \right)
Focus: (0,128)\left( {0, - \dfrac{1}{{28}}} \right)
Axis of Symmetry: x=0x = 0
Directrix: y=128y = \dfrac{1}{{28}}
Select a few xx values, and plug them into the equation to find the corresponding yy values. The xx values should be selected around the vertex.
Replace the variable xx with 1 - 1 in the expression.
f(1)=7(1)2f\left( { - 1} \right) = - 7{\left( { - 1} \right)^2}
Simplify the result.
f(1)=7f\left( { - 1} \right) = - 7
The final answer is 7 - 7.
The yy value at x=1x = - 1 is 7 - 7.
y=7y = - 7
Replace the variable xx with 2 - 2 in the expression.
f(2)=7(2)2f\left( { - 2} \right) = - 7{\left( { - 2} \right)^2}
Simplify the result.
f(2)=28f\left( { - 2} \right) = - 28
The final answer is 28 - 28.
The yy value at x=2x = - 2 is 28 - 28.
y=28y = - 28
Replace the variable xx with 11 in the expression.
f(1)=7(1)2f\left( 1 \right) = - 7{\left( 1 \right)^2}
Simplify the result.
f(1)=7f\left( 1 \right) = - 7
The final answer is 7 - 7.
The yy value at x=1x = 1 is 7 - 7.
y=7y = - 7
Replace the variable xx with 22 in the expression.
f(2)=7(2)2f\left( 2 \right) = - 7{\left( 2 \right)^2}
Simplify the result.
f(2)=28f\left( 2 \right) = - 28
The final answer is 28 - 28.
The yy value at x=2x = 2 is 28 - 28.
y=28y = - 28

xxyy
2 - 228 - 28
1 - 17 - 7
0000
117 - 7
2228 - 28

Graph the parabola using its properties and the selected points.
Direction: Opens Down
Vertex: (0,0)\left( {0,0} \right)
Focus: (0,128)\left( {0, - \dfrac{1}{{28}}} \right)
Axis of Symmetry: x=0x = 0
Directrix: y=128y = \dfrac{1}{{28}}

xxyy
2 - 228 - 28
1 - 17 - 7
0000
117 - 7
2228 - 28

Note: Graph of y=7x2y = 7{x^2} will open upwards with vertex at centre, focus at (0,128)\left( {0,\dfrac{1}{{28}}} \right), axis of symmetry x=0x = 0 and directrix y=128y = - \dfrac{1}{{28}}. It means a mirror image of y=7x2y = - 7{x^2}.