LEARNING TASK 1 Find the axis of symmetry, vertex and direction of the opening of the parabola. Axis of Vertex Opening symmetry of the parabola
Answers
Answer:
Please make brilliant answer
Step-by-step explanation:
. y=x^2+8x+4y=x
2
+8x+4
Opens upwards
Vertex: (-4, -12)(−4,−12)
Range: [-12, \infty)[−12,∞)
Axis of Symmetry: x=-12x=−12
x-intercept: (-4 \pm 2\sqrt{3}, 0)(−4±2
3
,0)
y-intercept: (0, 4)(0,4)
2. y = x^2 +4x + 5y=x
2
+4x+5
Opens upward
Vertex: (-2,1)(−2,1)
Range: [1, \infty)[1,∞)
Axis of Symmetry: x=-2x=−2
x-intercept: none
y-intercept: (0,5)(0,5)
3. y = -3x^2+ 12x - 7y=−3x
2
+12x−7
Opens downwards
Vertex: (2,5)(2,5)
Range: [5, -\infty)[5,−∞)
Axis of Symmetry: x=2x=2
x-intercept: \begin{aligned}\left(\frac{6 \pm \sqrt{15}}{3},0\right)\end{aligned}
(
3
6±
15
,0)
y-intercept: (0, -7)(0,−7)
4. y=2x^2-4x-6y=2x
2
−4x−6
Opens upwards
Vertex: (1, -8)(1,−8)
Range: [-8, \infty)[−8,∞)
Axis of Symmetry: x=-8x=−8
x-intercept: (3,0)(3,0) and (-1,0)(−1,0)
y-intercept: (0, -6)(0,−6)
Explanation
A Quadratic Function is in the form of F(x)=Ax^2+Bx+CF(x)=Ax
2
+Bx+C where A, B, C \in \mathbb{R}A,B,C∈R and A \ne 0A
=0 . This also means that the degree (the highest exponent) of a quadratic function should be 2.
Definition of Terms
The domain of a function is the set of real numbers in which the function is defined. Or this is the set of all possible xx values.
The range of a function is the set of corresponding output values for a given set of domain values. Or this is the set of all possible yy values.
The axis of symmetry is the line that crosses the vertex of the parabola and divides the parabola into two equal regions.
The xx -intercept of the function is where the function crosses or touches the x-axis.
The yy -intercept of the function is where the function crosses or touches the y-axis.
Solutions
The domain of all the functions given is the set of all real numbers or in notation, x \in (-\infty,\infty)x∈(−∞,∞) or \{x|x \in \mathbb{R}\}{x∣x∈R} . This is because all of the functions are defined under any real numbers.
1. y=x^2+8x+4y=x
2
+8x+4
Suppose a quadratic function in the form of F(x)=Ax^2+Bx+CF(x)=Ax
2
+Bx+C , then -A−A the function opens downwards and AA the function opens upward.
So the function y=f(x)=x^2+8x+4y=f(x)=x
2
+8x+4 opens upward, this means to say that its range is from the vertex then it goes up. Since the range starts from the yy element of vertex,\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)(−
2a
b
,f(−
2a
b
)) , then the range should be [f(-\frac{b}{2a} ), \infty)[f(−
2a
b
),∞) .
Substitute the values of a=1a=1 and b=8b=8 into f(-\frac{b}{2a})f(−
2a
b
) , then evaluate.
\begin{gathered}\begin{aligned}f\left(-\frac{b}{2a}\right)=f\left(-\frac{8}{2(1)}\right)&=f(-4)\\f(-4)&=(-4)^2+8(-4)+4\\&=16-32+4\\&=-12\end{aligned}\end{gathered}
f(−
2a
b
)=f(−
2(1)
8
)
f(−4)
=f(−4)
=(−4)
2
+8(−4)+4
=16−32+4
=−12
Thus, the range of the function is [-12, \infty)[−12,∞) .
This follows that the axis of symmetry should be
\begin{gathered}\begin{aligned}x&=-\frac{b}{2a}\\&=-\frac{8}{2(1)}\\&=-4\end{aligned}\end{gathered}
x
=−
2a
b
=−
2(1)
8
=−4
.
The vertex should be (-4,-12)(−4,−12) .
To solve for the intercepts, set x=0x=0 to solve for y-intercept and set y=0y=0 to solve for the x-intercept.
\begin{gathered}\begin{aligned}x&=0\\y&=(0)^2+8(0)+4\\&=4\\ &\therefore (0, 4) \;\text{is the y-intercept}\end{aligned}\end{gathered}
x
y
=0
=(0)
2
+8(0)+4
=4
∴(0,4)is the y-intercept
\begin{gathered}\begin{aligned}y&=0\\0&=x^2+8x+4\\-4\pm2\sqrt{3}&=x\:\text{use the quadratic formula}\\ &\therefore (-4+2\sqrt{3},0)\text{ and } (-4-2\sqrt{3}, 0) \;\text{are the x-intercepts}\end{aligned}\end{gathered}
y
0
−4±2
3
=0
=x
2
+8x+4
=xuse the quadratic formula
∴(−4+2
3
,0) and (−4−2
3
,0)are the x-intercepts
The answers for each item are achieved using the steps in item number 1.
2. y = x^2 +4x + 5y=x
2
+4x+5
Opens upward
Vertex: (-2,1)(−2,1)
Range: [1, \infty)[1,∞)
Axis of Symmetry: x=-2x=−2
x-intercept: none
y-intercept: (0,5)(0,5)
3. y = -3x^2+ 12x - 7y=−3x
2
+12x−7
Opens downwards
Vertex: (2,5)(2,5)
Range: [5, -\infty)[5,−∞)
Axis of Symmetry: x=2x=2
x-intercept: \begin{aligned}\left(\frac{6 \pm \sqrt{15}}{3},0\right)\end{aligned}
(
3
6±
15
,0)
y-intercept: (0, -7)(0,−7)
4. y=2x^2-4x-6y=2x
2
−4x−6
Opens upwards
Vertex: (1, -8)(1,−8)
Range: [-8, \infty)[−8,∞)
Axis of Symmetry: x=-8x=−8
x-intercept: (3,0)(3,0) and (-1,0)(−1,0)
y-intercept: (0, -6)(0,−6)
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To learn more about quadratic functions, go to
Zeros of a function: https://brainly.ph/question/2436149
Quadratic Formula: https://brainly.ph/question/2798405
Domain and Range: https://brainly.ph/question/2172206
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