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NO LINKS!! PART 3. Please help me with these graphs​

NO LINKS!! PART 3. Please help me with these graphs​-example-1

2 Answers

11 votes

#1

  • Vertex at (-5,4)

Equation

  • y=a(x+5)²+4

As there is no y inetercept present and the function is same as general quadratic equation y=x^2 just change in coordinates a=1

Final equation

  • y=(x+5)²+4

#2

  • Vertex (2,-3)

y=a(x-2)²-3

Use (4,-5)

  • -4=a(4-2)²-3
  • -1=a(2)²
  • 4a=-1
  • a=-1/4

Equation

  • y=-1/4(x-2)²-3
User MER
by
8.6k points
8 votes

Answer:


\textsf{13)} \quad y=(x+5)^2+4


\textsf{14)} \quad y=-(1)/(2)(x-2)^2-3

Explanation:

Vertex form of a parabola:


y=a(x-h)^2+k where (h, k) is the vertex

Question 13

From inspection of the graph, the vertex is (-5, 4)


\implies y=a(x+5)^2+4

To find
a, substitute the coordinates of a point on the curve into the equation.

Using point (-4, 5):


\implies a(-4+5)^2+4=5


\implies a(1)^2+4=5


\implies a+4=5


\implies a=1

Therefore, the equation of the parabola in vertex form is:


y=(x+5)^2+4

Question 14

From inspection of the graph, the vertex is (2, -3)


\implies y=a(x-2)^2-3

To find
a, substitute the coordinates of a point on the curve into the equation.

Using point (0, -5):


\implies a(0-2)^2-3=-5


\implies a(-2)^2-3=-5


\implies 4a-3=-5


\implies 4a=-2


\implies a=-(1)/(2)

Therefore, the equation of the parabola in vertex form is:


\implies y=-(1)/(2)(x-2)^2-3

User Draykos
by
7.8k points

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