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Inga [223]
2 years ago
14

Can someone do this fast please it’s like 2am

Mathematics
1 answer:
Lilit [14]2 years ago
5 0

Answer:

(5, 6)

Step-by-step explanation:

First, get both equations into slope-intercept form so they are easier to graph:

y=mx+b

The first equation is already in this form, so only do the second equation:

x-y=-1

To do this, just solve for y and that should almost always get it in the correct form.

\rightarrow x-y=-1\\\rightarrow -y=-x-1\\\rightarrow y=x+1

That equation is also now in slope-intercept form.

Now, using the slope and y-intercept in both equations, you can graph them. Using the graph, you can see the solution of this system of equations at (5, 6).

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A quadratic function has the form
y = ax^2 + bx + c

Use point (3, 5) in the equation above:

5 = a(3^2) + 3b + c
5 = 9a + 3b + c
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Use point (4, 3) in the equation above:

3 = a(4^2) + 4b + c
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Use point (5, 3) in the equation above.

5 = a(5^2) + 5b + c
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Now solve the system of equations of equations 1, 2, and 3 to find the coefficients, a, b, and c.

9a + 3b + c = 5
16a + 4b + c = 3
25a + 5b + c = 5

Subtract the first equation from the second equation.
Subtract the second equation from the third equation.
You get
7a + b = -2
9a + b = 2

Subtract the first equation above from the second equation to get.
2a = 4
a = 2

Substitute:
7a + b = -2
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b = -16

9a + 3b + c = 5
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The equation in standard form is

y = 2x^2 - 16x + 35

We can find it in vertex form:

y = 2(x^2 - 8x) + 35

y = 2(x^2 - 8x + 16) + 35 - 32

y = 2(x - 4)^2 + 3
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