Answer:
5 terms
to the fourth degree
leading coeff of 1
3 turning points
end behavior (when x -> inf, y -> inf. When x -> - inf, y -> -inf)
x intercepts are (0,-4) (0,-2) (0,1) (0,3)
Relative min: (-3.193, -25) (2.193, 25)
Relative max: (-0.5, 27.563)
Step-by-step explanation:
The terms can be counted, seperated by the + and - in the equation given.
The highest exponent is your degree.
The number before the highest term is your leading coeff, if there is no number it is 1.
The turning points are where the graph goes from falling to increasing or vice versa.
End behaviour you have to look at what why does when x goes to -inf and inf.
X int are the points at which the graph crosses the x-axis.
The relative min and max are findable if you plug in the graph on desmos or a graphing calculator.
Answer:
The correct option is B.
Step-by-step explanation:
Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .
Draw a diagonal AC.
In triangle BCA and DAC,
AC\cong ACAC≅AC (Reflexive Property of Equality)
\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)
\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)
The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.
By ASA postulate,
\triangle BCA\cong \triangle DAC△BCA≅△DAC
Therefore option B is correct
The answer is d, just find the equation and then plug in the first few numbers of copies and you’ll find the correct answer
mark me brainliest if the answer is correct, thank you in advance :)
Answer:
c. and Sagot po PA brainless po pleas
Answer:
a. Narrower
b. Shifts left
c. Opens up
d. Shifts up
Step-by-step explanation:
The original quadratic equation is y = x²
The given quadratic equation is y = 5·(x + 4)² + 7
The given quadratic equation is of the form, f(x) = a·(x - h)² + k
a. A quadratic equation is narrower than the standard form when the coefficient is larger than the coefficient in the original equation
The quadratic coefficient is 5 > 1 in the original, therefore, the quadratic equation is <em>narrower</em>
b. Given that the given quadratic equation has positive 'a', and 'b', and h = -4, therefore, the axis of symmetry <em>shifts left</em>
c. The quadratic coefficient is positive, (a = 5), therefore, the quadratic equation <em>opens down</em>
d. The value of 'k' gives the vertical shift, therefore, the given quadratic equation with k = 7, <em>shifts up.</em>
