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Match the graph with it's corresponding equation (1 point)

O y = −2x² + 2
O y=2x²
O y = 2x² + 2
O y = -x²​

Match the graph with it's corresponding equation (1 point) O y = −2x² + 2 O y=2x² O-example-1
User Trusktr
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The graph matches with (y = -2x²+ 2), depicting a downward-opening narrow parabola due to negative and absolute value greater than one coefficient for x² term and has its vertex shifted up because of positive constant.

The correct answer is "y = −2x² + 2".

The graph depicted in the image is a downward-opening parabola with its vertex above the origin, indicating a negative coefficient for the
(x^2) term and a positive y-intercept. This aligns with the equation
(y = -2x^2 + 2). In this equation, the coefficient of
(x^2) is -2, which ensures that the parabola opens downwards. The constant term +2 shifts the vertex of the parabola upwards by two units on the y-axis.

In quadratic equations like these, where they are expressed in the form
(y = ax^2 + bx + c), ‘a’ determines the direction in which the parabola opens. If ‘a’ is positive, it opens upwards; if negative, it opens downwards. The value of ‘c’ gives us our y-intercept – where our graph crosses or touches upon (in case of touching it would be at vertex) – and here it’s at (0, 2).

The steepness or width of this particular parabola is determined by an absolute value of ‘a’ being greater than one; hence making it narrower than standard unitary parabolas represented by equations like (y=x²). This specific characteristic can be observed from how quickly values on y decrease as x moves away from origin.

User Joe Minichino
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