one million) Foil (x ? 5)(2x + 3) so as which you have it interior the kind y=ax^2 + bx + c. you're able to get 2x^2 -7x -15, wherein case b= - 7 2) It would desire to have x^2 OR y^2 as a fashion to be a quadratic (yet you not often use the y^2 kind, so do no longer complication approximately it). "a" would not have the two. In "b" the x^2 cancels out once you simplify. In "c" you get x^3 after foiling that's a cubic function no longer a quadratic. "d" is the only decision the place you're left with an x^2 after simplifying. 3) while a quadratic is interior the kind of y=ax^2 + bx + c, you may tell if it opens up or down via staring on the verify in the "a" value. thus, a is a favorable selection and therefor the parabola opens up. it is going to appear like the letter "U." in case you seem at it, you will discover that it has a minimum value. If it had opened down, it might have a max. If I had extra time, i'd answer the rest. stable luck!
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Verified answer
y = 3x^2 + 18x + 23 = 3(x^2 + 6x + 9) - 4 = 3(x + 3)^2 - 4
y = 3x² + 18x + 23
factor out the leading coefficient
y = 3(x² + 6x) + 23
complete the square:
coefficient of x term is 6
half of coefficient is 3
use 3² to complete the square
y = 3(x² + 6x + 3²) - 3·3² + 23 note that you have to subtract 3·3² to keep the equation balanced
y = 3(x + 3)² - 4
vertex (-3, -4)
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y = 3x² + 18x + 23
y = 3(x² + 6x + _) + 23
y = 3(x² + 6x + 9) + 23 - 27
y = 3(x + 3)² - 4
Just complete the square y = 3( x^2 + 6x) + 23 = 3((x +3)^2- 9) + 23 = 3(x + 3)^2 - 4.
y = 3x² + 18x + 23
Find the x value of vertex, -b/2a
x = -3
Substitute it in the original equation.
y = -4
Vertex form is y = a(x - h)² + k [Where h,k is the vertex and a is the LEADING coefficient in the ORIGINAL equation]
Final
y = 3(x + 3)² - 4
one million) Foil (x ? 5)(2x + 3) so as which you have it interior the kind y=ax^2 + bx + c. you're able to get 2x^2 -7x -15, wherein case b= - 7 2) It would desire to have x^2 OR y^2 as a fashion to be a quadratic (yet you not often use the y^2 kind, so do no longer complication approximately it). "a" would not have the two. In "b" the x^2 cancels out once you simplify. In "c" you get x^3 after foiling that's a cubic function no longer a quadratic. "d" is the only decision the place you're left with an x^2 after simplifying. 3) while a quadratic is interior the kind of y=ax^2 + bx + c, you may tell if it opens up or down via staring on the verify in the "a" value. thus, a is a favorable selection and therefor the parabola opens up. it is going to appear like the letter "U." in case you seem at it, you will discover that it has a minimum value. If it had opened down, it might have a max. If I had extra time, i'd answer the rest. stable luck!