Question

the graph shows the equation $y = x^2 + 4x + 3$. based on the graph, describe the solutions of the equation. use the drop-down menus to complete the statements.

click the arrows to choose an answer from each menu.

the graph shows that there are no real solutions of the equation $y = x^2 + 4x + 3$ for values where choose... . the set of solutions of the equation includes choose...

Explanation:

Step1: Analyze the parabola's roots

The equation \( y = x^2 + 4x + 3 \) is a quadratic. Factoring, we get \( y=(x + 1)(x + 3) \), so the roots (where \( y = 0 \)) are \( x=-3 \) and \( x=-1 \). The parabola opens upward (since the coefficient of \( x^2 \) is positive). The vertex is at \( x = -\frac{b}{2a}=-\frac{4}{2(1)}=-2 \), and \( y=(-2)^2 + 4(-2)+3=-1 \). So the parabola is above the \( x \)-axis except between its two roots? No, wait: the vertex is at \( (-2, -1) \), so the parabola dips below the \( x \)-axis? Wait, no, \( y=-1 \) at \( x=-2 \), so the parabola has a minimum at \( y = -1 \), so it crosses the \( x \)-axis at \( x=-3 \) and \( x=-1 \). Wait, the graph in the image: let's check the \( x \)-intercepts. From the graph, the parabola crosses the \( x \)-axis at \( x=-3 \) and \( x=-1 \) (since at \( x=-3 \), \( y=0 \); at \( x=-1 \), \( y=0 \)). The vertex is at \( (-2, -1) \), so the parabola is below the \( x \)-axis between \( x=-3 \) and \( x=-1 \), and above otherwise. Wait, but the question is about "no real solutions" for the equation \( y = x^2 + 4x + 3 \) – wait, maybe the equation is \( x^2 + 4x + 3 = 0 \)? Wait, the original question says "the solutions of the equation" – maybe it's \( x^2 + 4x + 3 = 0 \), but the graph is of \( y = x^2 + 4x + 3 \). The solutions to \( x^2 + 4x + 3 = 0 \) are the \( x \)-intercepts (\( x=-3, x=-1 \)). But the first part: "there are no real solutions of the equation \( y = x^2 + 4x + 3 \) for values where..." Wait, maybe the equation is \( x^2 + 4x + 3 = k \) (some constant), so we're looking at when \( x^2 + 4x + 3 = k \) has no real solutions. For a quadratic \( ax^2+bx+c = k \), which is \( ax^2+bx+(c - k)=0 \), the discriminant is \( b^2 - 4a(c - k) \). For no real solutions, discriminant \( < 0 \). Here, \( a = 1 \), \( b = 4 \), \( c = 3 \), so discriminant \( = 16 - 4(1)(3 - k)=16 - 12 + 4k = 4 + 4k \). Set \( 4 + 4k < 0 \implies k < -1 \). So the equation \( x^2 + 4x + 3 = k \) has no real solutions when \( k < -1 \) (i.e., \( y < -1 \), since \( y = k \)). Wait, but the graph: the vertex is at \( y=-1 \), so the minimum value of \( y \) is \( -1 \). So the equation \( y = x^2 + 4x + 3 \) (where \( y \) is a constant) has no real solutions when \( y < -1 \) (because the parabola never goes below \( y = -1 \)). Then, the set of solutions (for \( y = x^2 + 4x + 3 = 0 \), wait no, the equation's solutions: if it's \( x^2 + 4x + 3 = 0 \), solutions are \( x=-3 \) and \( x=-1 \). But maybe the first drop-down is about \( y \) values: "no real solutions for values where \( y < -1 \)" (since the parabola's minimum is \( y=-1 \), so for \( y < -1 \), there's no \( x \) such that \( x^2 + 4x + 3 = y \)). Then the set of solutions (for the equation, maybe \( x^2 + 4x + 3 = 0 \)) includes \( x=-3 \) and \( x=-1 \). Wait, let's re-express:

The equation \( y = x^2 + 4x + 3 \): to find real solutions for \( x \) given \( y \), we solve \( x^2 + 4x + (3 - y)=0 \). The discriminant is \( 16 - 4(3 - y)=4 + 4y \). For real solutions, discriminant \( \geq 0 \implies 4 + 4y \geq 0 \implies y \geq -1 \). So when \( y < -1 \), there are no real solutions for \( x \). So the first blank: "no real solutions for values where \( y < -1 \)". The set of solutions (for the equation, maybe the roots of \( x^2 + 4x + 3 = 0 \)) are \( x=-3 \) and \( x=-1 \), so the solutions include \( x=-3 \) and \( x=-1 \).

Step2: Confirm with the graph

Looking at the graph, the parabola has a minimum at \( y=-1 \) (vertex at \( (-2, -1) \)) and crosses the \( x \)-axis at \( x=-3 \) and \( x=-1 \) (where \( y=0 \)). So for \( y < -1 \), there's no \( x \) (real) that satisfies \( y = x^2 + 4x + 3 \). The solutions to \( x^2 + 4x + 3 = 0 \) (if that's the equation) are \( x=-3 \) and \( x=-1 \), so the set of solutions includes those \( x \)-values.

Answer:

First drop-down: \( y < -1 \) (or the option corresponding to \( y \) less than -1). Second drop-down: the \( x \)-values \( -3 \) and \( -1 \) (or the option with \( x = -3 \) and \( x = -1 \)).

(Note: Since the drop-down options aren't listed, but based on the analysis, the first part is about \( y < -1 \) (no real solutions when \( y \) is less than the vertex's \( y \)-value), and the solutions are the \( x \)-intercepts \( x = -3 \) and \( x = -1 \).)