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3 years ago

A quadratic equation function f(x)=ax^2 passes through (0,0) and (5,5). write an equation for this function

1 answer:

Alex3 years ago

6 0

Equation: f(x)=(1/5)x^2 or f(x)=(0.2)x^2

Check:
f(0)=(1/5)(0)^2
f(0)=(1/5)(0)
f(0)=0✓
f(5)=(1/5)(5)^2
f(5)=(1/5)(25)
f(5)=5✓

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Solve 0 = (x – 4)2 – 1 by graphing the related function.

fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

$y=\left(x-\:4\right)^2-1$

Finding the x-intercept by substituting the value y = 0

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

$x=5$

$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

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I suspect you may have slipped up as you copied the question. Exactly as you wrote it, it has a single zero, at x= 2 .

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