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Find the quadratic equation for y=4x^2-5

2 Answers

2 votes

Answer:

y=4/x^3

Explanation:

User Joserobleda
by
5.3k points
2 votes

1) if you want to find the roots:

For this case we have the following quadratic equation:


y = 4x ^ 2-5

To find the solutions we do
y = 0:


4x ^ 2-5 = 0

We add 5 to both sides of the equation:


4x ^ 2 = 5

DIviding between 4 to both sides of the equation:


x ^ 2 = \frac {5} {4}

We apply square root to both sides:


x = \pm \sqrt {\frac {5} {4}}\\x = \pm \frac {\sqrt {5}} {2}

Thus, the roots are:


x_ {1} = \frac {\sqrt {5}} {2}\\x_ {2} = - \frac {\sqrt {5}} {2}

Answer:


x_ {1} = \frac {\sqrt {5}} {2}\\x_ {2} = - \frac {\sqrt {5}} {2}

2): if you want to write the equation in vertex form:

The general quadratic equation is:


y = a(x-h)^2+k

where,

a: is the leading coefficient

(h,k): is the verex of the quadratic equation

Comparing with the original equation we have


y = 4x ^ 2-5

So, the vertex is:


(h,k)=(0,-5)

Answer

The quadratic equation in vertex form is:


y = 4x ^ 2-5

User Mjmuk
by
5.0k points