Respuesta :

Simplify

[tex]\\ \rm\longmapsto y=(x-20)^2+5[/tex]

[tex]\\ \rm\longmapsto y=x^2-40x+405[/tex]

Graph attached

  • No real zeros
Ver imagen Аноним
Nayefx

Answer:

no real zeros

Step-by-step explanation:

We are given the following quadratic function:

[tex]y = {(x - 20)}^{2} + 5[/tex]

firstly simplify it to standard form which is

[tex]y = {x}^{2} -40x + 405[/tex]

To figure out the real zeros, set y to 0:

[tex] {x}^{2} - 40x + 405 = 0[/tex]

recall that,a quadratic equations has real zeros in case its discriminant is greater than or equal to 0, therefore

  • b²-4ac≥0⇒real roots

consider,

  • a=1
  • b=-40
  • c=405

now substitute:

[tex] { (- 40)}^{2} - 4(1)(405) \stackrel{ ? }{ \geq}0[/tex]

simplifying yields:

[tex] 1600 - 1620 \stackrel{ ? }{ \geq}0 \\ - 20 \ngeq 0[/tex]

hence, The function has no real zeros

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